Trigonometric Equations

$ \begin{aligned} & \sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi] \\ & \Rightarrow \quad \sqrt{3}\left(2 \cos ^2 \theta-1\right) 8 \cos \theta+3 \sqrt{3}=0 \\ & \Rightarrow \quad 2 \sqrt{3} \cos ^2 \theta+8 \cos \theta+2 \sqrt{3}=0 \\ & \Rightarrow \quad 2 \sqrt{3} \cos ^2 \theta+6 \cos \theta+2 \cos \theta+2 \sqrt{3}=0 \\ & \Rightarrow \quad 2 \sqrt{3} \cos \theta(\cos \theta+\sqrt{3})+2(\cos \theta+\sqrt{3})=0 \\ & \therefore \quad(2 \sqrt{3} \cos \theta+2)(\cos \theta+\sqrt{3})=0 \\ & \Rightarrow \cos \theta=\frac{-2}{2 \sqrt{3}} \\ & =-\frac{1}{\sqrt{3}} \end{aligned} $

Total 5 solutions

Option (4) is correct

(P)

$ \begin{aligned} & \sin ^6 x+\cos ^2 x=1, x \in[-\pi, \pi] \\ & \because \sin ^6 x \leq \sin ^2 x \text { and } \cos ^4 x \leq \cos ^2 x \text { for } x \in[-\pi, \pi] \\ & \therefore \sin ^6 x+\cos ^4 x \leq \sin ^2 x+\cos ^2 x-1 \end{aligned} $

Equality holds only when $\sin ^6 x=\sin ^2 x$ and $\cos ^4 x=\cos ^2 x$

This occurs when $\sin x \in\{-1,0,1\}$ and $\cos x \in\{-1,0,1\}$

$ \therefore x \in\left\{-\pi, \frac{-\pi}{2}, 0, \frac{\pi}{2}, \pi\right\} \Rightarrow 5 \text { solutions } \Rightarrow 5 \text { elements } $

(Q)

$ \begin{aligned} & \sin ^2 x+\cos ^6 x=1 \\ & 1-\cos ^2 x+\cos ^6 x=1 \\ & \cos ^6 x-\cos ^2 x=0 \\ & \cos ^2 x\left(\cos ^4 x-1\right)=0 \\ & \cos ^2 x\left(\cos ^2 x-1\right)\left(\cos ^2 x+1\right)=0 \\ & \because \cos ^2 x+1 \neq 0 \\ & \therefore \cos ^2 x=0 \Rightarrow x=\frac{\pi}{2}, \frac{-\pi}{2} \\ & \cos ^2 x=1 \Rightarrow x=0 \\ & \therefore 3 \text { elements. } \end{aligned} $

(R)

$ \begin{aligned} & \cos ^2 \frac{x}{2}-\sin ^2 x=\frac{1}{2} \\ & \frac{1+\cos x}{2}-\left(1-\cos ^2 x\right)=\frac{1}{2} \\ & \Rightarrow 1+\cos x-2+2 \cos ^2 x=1 \\ & 2 \cos ^2 x+\cos -2=0 \\ & \cos x=\frac{-1 \pm \sqrt{1+16}}{4} \\ & \cos x=\frac{-1+\sqrt{17}}{4}, \frac{-1-\sqrt{17}}{4} \\ & \text { only possibility } \cos x=\frac{-1+\sqrt{17}}{4}, \frac{-1-\sqrt{17}}{4} \end{aligned} $

Number of elements in $[-\pi, \pi]$ is 2

(S)

$ \begin{aligned} & 6 \sin ^2 \frac{x}{2}-\cos 3 x=3, x \in[-2 \pi, 2 \pi] \\ & 3(1-\cos x)-\cos 3 x=3 \\ & \Rightarrow-3 \cos x-\cos 3 x=0 \\ & \Rightarrow \cos 3 x+3 \cos x=0 \\ & \Rightarrow 4 \cos ^3 x=0 \Rightarrow \cos x=0 \end{aligned} $

For $x \in[-2 \pi, 2 \pi]$

$ \cos x=0 \Rightarrow x \in\left\{\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}\right\} $

$\Rightarrow 4$ elements.

$ (P) \rightarrow(5),(Q) \rightarrow(3),(R) \rightarrow(2),(S) \rightarrow(4) $

Option (B) is correct.

Given,

$ \cos \theta + 1 = \sqrt{3}\sin \theta $

We have to find all $\theta \in (-2\pi, 2\pi)$ satisfying this equation, and then calculate their sum.

Rewrite the equation as

$ 1+\cos\theta=\sqrt{3}\sin\theta $

Now use the standard identities:

$ 1+\cos\theta = 2\cos^2\frac{\theta}{2} $

and

$ \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} $

Substituting these,

$ 2\cos^2\frac{\theta}{2}=\sqrt{3}\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} $

So,

$ 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}-\sqrt{3}\sin\frac{\theta}{2}\right)=0 $

Hence, either

Case 1:

$ \cos\frac{\theta}{2}=0 $

So,

$ \frac{\theta}{2}=\frac{\pi}{2}+n\pi $

which gives

$ \theta=\pi+2n\pi $

Now in the interval $(-2\pi,2\pi)$, the possible values are:

$ \theta=-\pi,\ \pi $

Case 2:

$ \cos\frac{\theta}{2}-\sqrt{3}\sin\frac{\theta}{2}=0 $

So,

$ \cos\frac{\theta}{2}=\sqrt{3}\sin\frac{\theta}{2} $

or

$ \tan\frac{\theta}{2}=\frac{1}{\sqrt{3}} $

Thus,

$ \frac{\theta}{2}=\frac{\pi}{6}+n\pi $

So,

$ \theta=\frac{\pi}{3}+2n\pi $

Now in the interval $(-2\pi,2\pi)$, the possible values are:

$ \theta=-\frac{5\pi}{3},\ \frac{\pi}{3} $

Therefore,

$ S=\left\{-\pi,\ \pi,\ -\frac{5\pi}{3},\ \frac{\pi}{3}\right\} $

Now sum these values:

$ -\pi+\pi-\frac{5\pi}{3}+\frac{\pi}{3} $

$ = -\frac{4\pi}{3} $

Hence,

$ \sum_{\theta\in S}\theta = -\frac{4\pi}{3} $

So the correct answer is:

$ \boxed{-\frac{4\pi}{3}} $

Option B

Given

$ 3\sin^2 x + 12\cos x - 3 = p,\qquad x\in \mathbb{R} $

We need all integral values of $p$ for which this equation has at least one real solution.

Now use the identity

$ \sin^2 x = 1 - \cos^2 x $

So,

$ 3(1-\cos^2 x)+12\cos x-3 = p $

Simplifying,

$ 3 - 3\cos^2 x + 12\cos x - 3 = p $

$ -3\cos^2 x + 12\cos x = p $

Let

$ t = \cos x $

Then

$ -1 \le t \le 1 $

So we have

$ p = -3t^2 + 12t $

Now we need the range of

$ f(t) = -3t^2 + 12t,\qquad -1 \le t \le 1 $

This is a downward opening parabola.

Find its values at the endpoints:

$ f(-1) = -3(1) + 12(-1) = -3 - 12 = -15 $

$ f(1) = -3(1) + 12(1) = -3 + 12 = 9 $

Also,

$ f'(t) = -6t + 12 $

For $-1 \le t \le 1$, we have

$ -6t + 12 > 0 $

so $f(t)$ is increasing on $[-1,1]$.

Hence the range of $p$ is

$ -15 \le p \le 9 $

Therefore all integral values of $p$ are

$ -15,-14,-13,\dots,9 $

This is an AP with first term $-15$, last term $9$.

Number of terms:

$ n = 9-(-15)+1 = 25 $

Sum:

$ S = \frac{n}{2}(a+l) = \frac{25}{2}(-15+9) = \frac{25}{2}(-6) = -75 $

Hence, the required sum is

$ \boxed{-75} $

So, the correct option is:

$ \boxed{\text{Option C}} $

We need to find the number of elements in

$ S=\{x\in[-\pi,\pi]: \sin x(\sin x+\cos x)=a,\; a\in\mathbb Z\}. $

So we must find all $x\in[-\pi,\pi]$ for which the value

$ \sin x(\sin x+\cos x) $

is an integer.

Step 1: Simplify the expression

Let

$ f(x)=\sin x(\sin x+\cos x). $

Then

$ f(x)=\sin^2 x+\sin x\cos x. $

Now use the identities:

$ \sin^2 x=\frac{1-\cos 2x}{2}, \qquad \sin x\cos x=\frac{\sin 2x}{2}. $

So,

$ f(x)=\frac{1-\cos 2x}{2}+\frac{\sin 2x}{2} = \frac{1+\sin 2x-\cos 2x}{2}. $

Also,

$ \sin 2x-\cos 2x=\sqrt{2}\sin\left(2x-\frac{\pi}{4}\right). $

Hence,

$ f(x)=\frac{1+\sqrt{2}\sin\left(2x-\frac{\pi}{4}\right)}{2}. $

Step 2: Find the range of $f(x)$

Since

$ -1\le \sin\left(2x-\frac{\pi}{4}\right)\le 1, $

we get

$ -\sqrt{2}\le \sqrt{2}\sin\left(2x-\frac{\pi}{4}\right)\le \sqrt{2}. $

Therefore,

$ \frac{1-\sqrt{2}}{2}\le f(x)\le \frac{1+\sqrt{2}}{2}. $

Now,

$ \frac{1-\sqrt{2}}{2}\approx -0.207,\qquad \frac{1+\sqrt{2}}{2}\approx 1.207. $

So the only integers possible in this interval are

$ 0 \quad \text{and} \quad 1. $

Thus we need to solve:

1. $f(x)=0$

2. $f(x)=1$

Step 3: Solve $f(x)=0$

We have

$ \sin x(\sin x+\cos x)=0. $

So either

Case 1:

$ \sin x=0 $

In $[-\pi,\pi]$, this gives

$ x=-\pi,\;0,\;\pi. $

Case 2:

$ \sin x+\cos x=0 $

That is,

$ \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=0 $

so

$ x+\frac{\pi}{4}=n\pi \quad\Rightarrow\quad x=n\pi-\frac{\pi}{4}. $

Now check in $[-\pi,\pi]$:

• for $n=0$, $x=-\frac{\pi}{4}$

• for $n=1$, $x=\frac{3\pi}{4}$

• other values go outside the interval

So from $f(x)=0$, we get

$ x=-\pi,\;0,\;\pi,\;-\frac{\pi}{4},\;\frac{3\pi}{4}. $

Total $5$ values.

Step 4: Solve $f(x)=1$

We need

$ \sin x(\sin x+\cos x)=1. $

Using the simplified form:

$ \frac{1+\sqrt{2}\sin\left(2x-\frac{\pi}{4}\right)}{2}=1. $

So,

$ 1+\sqrt{2}\sin\left(2x-\frac{\pi}{4}\right)=2 $

$ \sqrt{2}\sin\left(2x-\frac{\pi}{4}\right)=1 $

$ \sin\left(2x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}. $

Now,

$ \sin \theta=\frac{1}{\sqrt{2}} $

when

$ \theta=\frac{\pi}{4}+2n\pi \quad\text{or}\quad \theta=\frac{3\pi}{4}+2n\pi. $

So,

First possibility:

$ 2x-\frac{\pi}{4}=\frac{\pi}{4}+2n\pi $

$ 2x=\frac{\pi}{2}+2n\pi $

$ x=\frac{\pi}{4}+n\pi $

In $[-\pi,\pi]$, this gives

$ x=\frac{\pi}{4},\;-\frac{3\pi}{4}. $

Second possibility:

$ 2x-\frac{\pi}{4}=\frac{3\pi}{4}+2n\pi $

$ 2x=\pi+2n\pi $

$ x=\frac{\pi}{2}+n\pi $

In $[-\pi,\pi]$, this gives

$ x=\frac{\pi}{2},\;-\frac{\pi}{2}. $

So from $f(x)=1$, we get $4$ values:

$ x=\frac{\pi}{4},\;-\frac{3\pi}{4},\;\frac{\pi}{2},\;-\frac{\pi}{2}. $

Step 5: Total number of elements

Combining both cases:

• from $a=0$: $5$ values

• from $a=1$: $4$ values

Hence,

$ n(S)=5+4=9. $

So the correct answer is:

$ \boxed{9} $

Therefore, Option D is correct.

$\begin{aligned} &\begin{aligned} & \cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \\ & \frac{1}{2}\left(2 \cos 2 \theta \cos \frac{\theta}{2}\right)+\cos \frac{50}{2} \\ & =\frac{1}{2}\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right) \end{aligned}\\ &\text { or solving }\\ &\begin{aligned} & \cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2} \\ & \cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}=0 \\ & 2 \sin 30 \sin \frac{9 \theta}{2}=0 \end{aligned}\\ &3 \theta=\mathrm{n} \pi \quad \text { or } \frac{9 \theta}{2}=\mathrm{m} \pi\\ &\begin{aligned} & \theta=\frac{\mathrm{n} \pi}{3} \quad \theta=\frac{2 \mathrm{~m} \pi}{9} \\ & \theta=\left\{-\frac{\pi}{2}, \frac{\pi}{3}, 0\right\} \\ & \theta=\left\{-\frac{4 \pi}{9}, \frac{-2 \pi}{9}, \frac{4 \pi}{9}, \frac{2 \pi}{9}\right\} \end{aligned} \end{aligned}$

To find the number of solutions to the equation

$ (4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, \quad x \in \left[-2\pi, \frac{5\pi}{2}\right] $

we start by letting $\sin x = t$, which implies $\cos^2 x = 1 - t^2$. Substituting these into the equation, we have:

$ (4-\sqrt{3}) t - 2 \sqrt{3}(1-t^2) = \frac{-4}{1+\sqrt{3}} $

Simplifying, we get:

$ 2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2\sqrt{3} + \frac{4}{1+\sqrt{3}} = 0 $

Further simplification yields:

$ 2 \sqrt{3} t^2 + (4-\sqrt{3}) t + \frac{-2 \sqrt{3} - 2}{1+\sqrt{3}} = 0 $

This simplifies to:

$ 2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2 = 0 $

To find $t$, solve the quadratic equation:

$ t = \frac{(\sqrt{3}-4) \pm \sqrt{19 - 8 \sqrt{3} + 8(2\sqrt{3})}}{4 \sqrt{3}} $

This simplifies to:

$ t = \frac{(\sqrt{3}-4) \pm \sqrt{19 + 8 \sqrt{3}}}{4 \sqrt{3}} = \frac{(\sqrt{3}-4) \pm (\sqrt{3}+4)}{4 \sqrt{3}} $

Evaluating further, we find:

$ t = \frac{2 \sqrt{3}}{4 \sqrt{3}} \quad \text{or} \quad \frac{-8}{4 \sqrt{3}} $

This gives us $ \sin x = \frac{1}{2} $ or $\frac{-2}{\sqrt{3}} < -1$. Since $\frac{-2}{\sqrt{3}}$ is less than -1, it is not valid for $\sin x$. Hence, the only solution is $\sin x = \frac{1}{2}$.

Considering the interval $x \in \left[-2\pi, \frac{5\pi}{2}\right]$, we find that $\sin x = \frac{1}{2}$ occurs at several points. After checking, there are 5 such solutions in the given interval.

$\begin{aligned} & 2 x+3 \tan x=\pi \\ & \Rightarrow \quad \tan x=\frac{\pi-2 x}{3}, x \in[-2 \pi, 2 \pi]-\left\{ \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}\right\} \end{aligned}$

5 solutions

To find the solutions of the equation $\sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0$, we can begin by solving for $\operatorname{cosec} \theta$.

The quadratic equation in terms of $\operatorname{cosec} \theta$ is:

$ \sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0 $

Using the quadratic formula:

$ \operatorname{cosec} \theta = \frac{2\sqrt{3} - 2 \pm \sqrt{(2(\sqrt{3} - 1))^2 + 4 \sqrt{3} \cdot 4}}{2\sqrt{3}} $

Simplifying inside the square root:

$ (2(\sqrt{3} - 1))^2 + 4\sqrt{3} \cdot 4 = 4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3} $

This simplifies to:

$ 4(4 - 2\sqrt{3}) + 16\sqrt{3} = 16 - 8\sqrt{3} + 16\sqrt{3} = 16 + 8\sqrt{3} $

Therefore, the quadratic gives:

$ \operatorname{cosec} \theta = \frac{2\sqrt{3} - 2 \pm 2(\sqrt{3} + 1)}{2\sqrt{3}} $

By solving, we find:

$ \operatorname{cosec} \theta = \frac{-2}{\sqrt{3}}, 2 $

Thus, for each value of the cosecant, $\theta$ can take several values that fall within the given interval $[- \frac{7\pi}{6}, \frac{4\pi}{3}]$:

$ \theta = \frac{-7\pi}{6}, \frac{-2\pi}{3}, \frac{-\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{4\pi}{3} $

These are all the possible solutions for $\theta$ within the defined range.

$\begin{aligned} & 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 \\ & 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\ & (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\ & \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text { or } \cos \theta=\frac{-1}{\sqrt{2}} \\ & \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\ & \Rightarrow 8 \text { (solution) } \end{aligned}$

$\begin{aligned} & 2 \sin ^2 \theta=\cos 2 \theta \\ & 2 \sin ^2 \theta=1-2 \sin ^2 \theta \\ & 4 \sin ^2 \theta=1 \\ & \sin ^2 \theta=\frac{1}{4} \\ & \sin \theta= \pm \frac{1}{2} \\ & 2 \cos ^2 \theta=3 \sin \theta \\ & 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & (2 \sin \theta-1)(2 \sin \theta-2)=0 \\ & \sin \theta=\frac{1}{2} \end{aligned}$

so common equation which satisfy both equations is $\sin \theta=\frac{1}{2}$

$\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])$

$\text { Sum }=\pi$

$\begin{aligned} & |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \\ & \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \\ & \cos 3 \theta \text { is max if } \cos 3 \theta=\frac{1}{2} \\ & \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \\ & \sum \theta_i=6 \pi \end{aligned}$

We start by recognizing that $ \sin^2 x + \cos^2 x = 1$. Substituting $ \sin^2 x = 1 - \cos^2 x$ into the original equation gives:

$ 4(1 - \cos^2 x) - 4 \cos^3 x + 9 - 4 \cos x = 0 $

Rearranging and simplifying this equation, we have:

$ 4 - 4 \cos^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$

$ 4 \cos^3 x + 4 \cos^2 x + 4 \cos x - 13 = 0$

$4 \cos ^3 x+4 \cos ^2 x+4 \cos x=13$

Observing the bounds given, $x \in [-2\pi, 2\pi]$, we want to find how many solutions satisfy this cubic equation in terms of $ \cos x$. However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when $ \cos x = 1$, that being $4(1) + 4(1) + 4(1) = 12$. Yet, we have the equation set to equal 13, which is impossible given the maximum sum of the LHS can only be 12.

The above reasoning indicates that, within the domain specified, there is no value of $x$ for which the equation holds true. Therefore, the number of solutions to the equation is zero.

$\begin{aligned} & 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\ & 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\ & 6 \sin x-4=0 \\ & \sin x=\frac{2}{3} \\ & \mathbf{n}=5 \text { (in the given interval) } \\ & x^2+5 x+2=0 \\ & x=\frac{-5 \pm \sqrt{17}}{2} \\ & \text { Required interval }(-\infty, 0) \end{aligned}$

$\begin{aligned} & \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\ & \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\ & x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0 \end{aligned}$

so, sum of real solutions $=-1$

$4+5 \tan \theta=\sec \theta$

Squaring : $24 \tan ^2 \theta+40 \tan \theta+15=0$

$\tan \theta=\frac{-10 \pm \sqrt{10}}{12}$

and $\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)$ is Rejected.

(3) is correct.

$\begin{aligned} & 2 \tan ^2 \theta-5 \sec \theta-1=0 \\ & \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\ & \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\ & \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\ & \Rightarrow \cos \theta=-2, \frac{1}{3} \\ & \Rightarrow \cos \theta=\frac{1}{3} \end{aligned}$

For 7 solutions $\mathrm{n}=13$

$\begin{aligned} & \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \\ & \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \\ & \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\ & \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}} \end{aligned}$

Given the information, let's start by analyzing the trigonometric relationships involving $\cot x = \frac{-5}{\sqrt{11}}$ where $\frac{\pi}{2} < x < \pi$.

We know that $\cot x$ is the reciprocal of $\tan x$. Hence,

$ \cot x = \frac{\cos x}{\sin x} = \frac{-5}{\sqrt{11}} $

From the above, it follows that:

$ \cos x = -5k \text{ and } \sin x = \sqrt{11}k $

for some constant $k$. Using the Pythagorean identity:

$ \cos^2 x + \sin^2 x = 1 $

Let’s substitute the values of $\cos x$ and $\sin x$ into this identity:

$ (-5k)^2 + (\sqrt{11}k)^2 = 1 $

$ 25k^2 + 11k^2 = 1 $

$ 36k^2 = 1 $

$ k^2 = \frac{1}{36} $

$ k = \frac{1}{6} \text{ or } k = -\frac{1}{6} $

Therefore, we have two sets of values:

$ \cos x = -\frac{5}{6} \text{ and } \sin x = \frac{\sqrt{11}}{6} $

Given that $x$ is in the interval $\left(\frac{\pi}{2}, \pi\right)$, where sine is positive and cosine is negative, we take:

$ \cos x = -\frac{5}{6},\ \sin x = \frac{\sqrt{11}}{6} $

Now consider the expression:

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x - \cos 6 x) + \left(\cos \frac{11 x}{2}\right)(\sin 6 x + \cos 6 x) $

To solve this, first find $\frac{11x}{2}$ which lies in the interval $\left(\frac{11\pi}{4}, \frac{11\pi}{2}\right)$. Next, we simplify each term using sum-to-product identities:

Notice:

$ \sin 6x - \cos 6x = \sqrt{2} \sin \left( 6x - \frac{\pi}{4} \right) $

$ \sin 6x + \cos 6x = \sqrt{2} \cos \left( 6x - \frac{\pi}{4} \right) $

Using these, the given expression becomes:

$ \left(\sin \frac{11x}{2}\right) \sqrt{2} \sin \left(6x - \frac{\pi}{4} \right) + \left(\cos \frac{11x}{2}\right) \sqrt{2} \cos \left(6x - \frac{\pi}{4} \right) $

This further simplifies using the angle sum identities:

$ = \sqrt{2} \left(\sin \frac{11x}{2} \sin \left(6x - \frac{\pi}{4}\right) + \cos \frac{11x}{2} \cos \left(6x - \frac{\pi}{4}\right)\right) $

Utilizing the cosine of the difference identity:

$ \sin A \sin B + \cos A \cos B = \cos(A - B) $

Substitute $A = \frac{11x}{2}$ and $B = 6x - \frac{\pi}{4}$:

$ \cos \left(\frac{11x}{2} - \left(6x - \frac{\pi}{4}\right)\right) = \cos \left(\frac{11x}{2} - 6x + \frac{\pi}{4}\right) = \cos \left( - \frac{x}{2} + \frac{\pi}{4} \right) = \cos \left(\frac{\pi}{4} - \frac{x}{2}\right) $

Using the values of trigonometric functions:

$ \cos \left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{\sqrt{11}+1}{2\sqrt{3}} $

Thus, the correct option is:

Option B

$ \frac{\sqrt{11}+1}{2 \sqrt{3}} $

The original equation is:

$ 3 \cos ^4 \theta-5 \cos ^2 \theta-2 \sin ^6 \theta+2=0 $

We can re-arrange terms and use trigonometric identity ($\cos^2 \theta + \sin^2 \theta = 1$) to rewrite this as :

$ 3 \cos ^4 \theta - 3 \cos ^2 \theta + 2 \sin ^2 \theta - 2 \sin ^6 \theta = 0 $

Factoring out $\cos^2 \theta$ and $\sin^2 \theta$ we get :

$ 3 \cos ^2 \theta (\cos ^2 \theta - 1) + 2 \sin ^2 \theta (\sin ^4 \theta - 1) = 0 $

Then we substitute $1 - \cos^2 \theta$ for $\sin^2 \theta$ (again using the same trigonometric identity), which simplifies to :

$ -3 \cos ^2 \theta \sin ^2 \theta + 2 \sin ^2 \theta(1 + \sin ^2 \theta)\cos ^2 \theta = 0 $

This can be re-written as :

$ \sin ^2 \theta \cos ^2 \theta(2 + 2 \sin ^2 \theta - 3) = 0 $

Simplifying it to :

$ \sin ^2 \theta \cos ^2 \theta(2 \sin ^2 \theta - 1) = 0 $

We now have 3 separate conditions to solve for :

- $\sin ^2 \theta = 0$,

- $\cos ^2 \theta = 0$, and

- $2 \sin ^2 \theta - 1 = 0$.

For $\sin ^2 \theta = 0$, the solutions are $\theta=\{0, \pi, 2 \pi\}$ (3 solutions).

For $\cos ^2 \theta = 0$, the solutions are $\theta=\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}$ (2 solutions).

For $2 \sin ^2 \theta - 1 = 0$ (which simplifies to $\sin ^2 \theta = 1/2$), the solutions are $\theta=\left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$ (4 solutions).

In total, this gives 3+2+4 = 9 solutions.

We have, $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$ and

$\beta=\sum_\limits{x \in S} \tan ^{2}\left(\frac{x}{3}\right)$

$ \begin{aligned} & 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10 \\\\ &\Rightarrow \frac{9}{9^{\tan ^2 x}}+9^{\tan ^2 x}=10 \end{aligned} $

Let $9^{\tan ^2 x}=t$

$ \frac{9}{t}+t=10 \Rightarrow t^2-10 t+9=0 \Rightarrow t=9 \text { or } 1 $

If $t=9,9^{\tan ^2 x}=9 \Rightarrow x= \pm \frac{\pi}{4}$

If $t=1,9^{\tan ^2 x}=1 \Rightarrow x=0$

Also,

$ \begin{aligned} \beta & =\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right) \\\\ & =\tan ^2\left(\frac{0}{3}\right)+\tan ^2\left(\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right) \\\\ & =0+2 \tan ^2 \frac{\pi}{12}=2[2-\sqrt{3}]^2 \\\\ & =2[4+3-4 \sqrt{3}]=14-8 \sqrt{3} \end{aligned} $

$ \text { So, now } \frac{1}{6}(\beta-14)^2=\frac{1}{6}[14-8 \sqrt{3}-14]^2=\frac{1}{6} \times 64 \times 3=32 $

Given,

$2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}$

We know,

A.M $\ge$ G.M

$\therefore$ for 4^x and 4^$-$x

${{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}} $

$ \Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1$

$ \Rightarrow {4^x} + {4^{ - x}} \ge 2$ ...... (1)

And we know,

$ - 1 \le \cos x \le 1$

$\therefore$ $ - 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1$

$ - 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2$ ...... (2)

(1) and (2) both satisfies only when both equal to 2.

$\therefore$ $2\cos \left( {{{{x^2} + x} \over 6}} \right) = 2$

$ \Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1$

$ \Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0$

$ \Rightarrow {{{x^2} + x} \over 6} = 0$

$ \Rightarrow {x^2} + x = 0$

$ \Rightarrow x(x + 1) = 0$

$ \Rightarrow x = 0, - 1$

When $x = 0$,

L.H.S $ = 2\cos \left( {{{{x^2} + x} \over 6}} \right)$

$ = 2\cos \left( {{0 \over 6}} \right)$

$ = 2\cos 0$

$ = 2\,.\,1$

$ = 2$

R.H.S $ = {4^x} + {4^{ - x}}$

$ = {4^0} + {4^0}$

$ = 2$

$\therefore$ $x = 0$ is accepted.

Now, when $x = - 1$,

L.H.S $ = 2\cos \left( {{{{x^2} + x} \over 6}} \right)$

$ = 2\cos \left( {{{1 - 1} \over 6}} \right)$

$ = 2\cos 0 = 2$

R.H.S $ = {4^x} + {4^{ - x}}$

$ = {4^{ - 1}} + {4^1}$

$ = {1 \over 4} + 4$

$ = {{17} \over 4}$

$\therefore$ L.H.S $\ne$ R.H.S

$\therefore$ $x = - 1$ is not a solution.

$\therefore$ Only one solution possible which is $x = 0$.

$s = \left\{ {0 \in \left( {0,{\pi \over 2}} \right):\sum\limits_{m = 1}^9 {\sec \left( {\theta + (m - 1){\pi \over 6}} \right)\sec \left( {\theta + {{m\pi } \over 6}} \right) = - {8 \over {\sqrt 3 }}} } \right\}$.

$\sum\limits_{m = 1}^9 {{1 \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)}}\cos \left( {\theta + m{\pi \over 6}} \right)} $

${1 \over {\sin \left( {{\pi \over 6}} \right)}}\sum\limits_{m = 1}^9 {{{\sin \left[ {\left( {\theta + {{m\pi } \over 6}} \right) - \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]} \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)\cos \left( {\theta + m{\pi \over 6}} \right)}}} $

$ = 2\sum\limits_{m = 1}^9 {\left[ {\tan \left( {\theta + {{m\pi } \over 6}} \right) - \tan \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]} $

Now,

$\matrix{ {m = 1} & {2\left[ {\tan \left( {\theta + {\pi \over 6}} \right) - \tan (\theta )} \right]} \cr {m = 2} & {2\left[ {\tan \left( {\theta + {{2\pi } \over 6}} \right) - \tan \left( {\theta + {\pi \over 6}} \right)} \right]} \cr {\matrix{ . \cr . \cr . \cr } } & {} \cr {m = 9} & {2\left[ {\tan \left( {\theta + {{9\pi } \over 6}} \right) - \tan \left( {\theta + 8{\pi \over 6}} \right)} \right]} \cr } $

$\therefore$ $ = 2\left[ {\tan \left( {\theta + {{3\pi } \over 2}} \right) - \tan \theta } \right] = {{ - 8} \over {\sqrt 3 }}$

$ = - 2[\cot \theta + \tan \theta ] = {{ - 8} \over {\sqrt 3 }}$

$ = - {{2 \times 2} \over {2\sin \theta \cos \theta }} = {{ - 8} \over {\sqrt 3 }}$

$ = {1 \over {\sin 2\theta }} = {2 \over {\sqrt 3 }}$

$ \Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}$

$2\theta = {\pi \over 3}$

$2\theta = {{2\pi } \over 3}$

$\theta = {\pi \over 6}$

$\theta = {\pi \over 3}$

$\sum {{\theta _i} = {\pi \over 6} + {\pi \over 3} = {\pi \over 2}} $

$S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }} = 16} \right\}$

Now apply AM $\ge$ GM for ${8^{2{{\sin }^2}\theta }},\,{8^{2{{\cos }^2}\theta }}$

${{{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }}} \over 2} \ge {\left( {{8^{2{{\sin }^2}\theta + 2{{\cos }^2}\theta }}} \right)^{{1 \over 2}}}$

$8 \ge 8$

$ \Rightarrow {8^{2{{\sin }^2}\theta }} = {8^{2{{\cos }^2}\theta }}$

or ${\sin ^2}\theta = {\cos ^2}\theta $

$\therefore$ $\theta = {\pi \over 4},{{3\pi } \over 4},{{5\pi } \over 4},{{7\pi } \over 4}$

$n(S) + \sum\limits_{\theta \in S}^{} {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} $

$4 + \sum\limits_{\theta \in S}^{} {{2 \over {2\sin \left( {{\pi \over 4} + 2\theta } \right)\cos \left( {{\pi \over 4} + 2\theta } \right)}}} $

$ = 4 + \sum\limits_{\theta \in S}^{} {{2 \over {\sin \left( {{\pi \over 2} + 4\theta } \right)}} = 4 + 2\sum\limits_{\theta \in S}^{} {\cos ec\left( {{\pi \over 2} + 4\theta } \right)} } $

$ = 4 + 2\left[ {\cos ec\left( {{\pi \over 2} + \pi } \right)\cos ec\left( {{\pi \over 2} + 3\pi } \right) + \cos ec\left( {{\pi \over 2} + 5\pi } \right) + \cos ec\left( {{\pi \over 2} + 7\pi } \right)} \right]$

$ = 4 + 2\left[ { - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2}} \right]$

$ = 4 - 2(4)$

$ = 4 - 8$

$ = - 4$

Period of $|\cos x| = \pi $

And period of $\sin x = 2\pi $

Graph of $\sin x$ and $|\cos x|$ cuts each other at two points A and B in $\left[ {0,2\pi } \right]$

So, in $\left[ { - 4\pi ,4\pi } \right]$, total 4 similar graph will be present and graph of $\sin x$ and $|\cos x|$ will cut $4 \times 2 = 8$ times.

$\therefore$ Total possible solutions = 8.

Given $a = \alpha - i\beta $ and

$4ix + (1 + i)y = 0$ ...... (i)

$8\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)x + \overline a y = 0$ .... (ii)

By (i)

${x \over y} = {{ - (1 + i)} \over {4i}}$ ...... (iii)

By (ii)

${x \over y} = {{ - \overline a } \over {8\left( {{{ - 1} \over 2} + {{\sqrt 3 i} \over 2}} \right)}}$ ..... (iv)

Now by (iii) and (iv)

${{1 + i} \over {4i}} = {{\overline a } \over {4\left( { - 1 + \sqrt 3 i} \right)}}$

$ \Rightarrow \overline a = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i$

$ \Rightarrow \alpha + i\beta = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i$

$\therefore$ ${\alpha \over \beta } = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3 $

$\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x,\,x \in [ - 3\pi ,\,3\pi ]$

$ \Rightarrow \cos 2x + \cos {{2\pi } \over 3} = {1 \over 2}{\cos ^2}2x$

$ \Rightarrow {\cos ^2}2x - 2\cos 2x - 1 = 0$

$ \Rightarrow \cos 2x = 1$

$\therefore$ $x = - 3\pi ,\, - 2\pi ,\, - \pi ,\,0,\,\pi ,\,2\pi ,\,3\pi $

$\therefore$ Number of solutions = 7

$\tan \theta(\sin \theta+1)-\sin 2 \theta=0$

$ \begin{aligned} &\tan \theta\left(\sin \theta+1-2 \cos ^{2} \theta\right)=0 \\\\ &\Rightarrow \tan \theta=0 \text { or } 2 \sin ^{2} \theta+\sin \theta-1=0 \\\\ &\Rightarrow(2 \sin \theta+1)(\sin \theta-1)=0 \\\\ &\Rightarrow \sin \theta=\frac{-1}{2} \text { or } 1 \end{aligned} $

But, $\sin \theta=1$ not possible

$ \theta=0, \pi,-\pi,-\frac{\pi}{6}, \frac{-5 \pi}{6} $

$ \mathrm{n}(\mathrm{S})=5 $

$ T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi) +\cos \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{\pi}{3}\right) $

= 4

Solving all question one by one we get,

$ \begin{aligned} \text { (i) } & \left\{x \in\left[\frac{-2 \pi}{3}, \frac{2 \pi}{3}\right], \cos x+\sin x=1\right\} \\\\ & \cos x+\sin x=1 \\\\ \Rightarrow \quad & \sin \left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}} \\\\ \Rightarrow & \frac{\pi}{4}+x=n \pi+(-1)^n \frac{\pi}{4} \end{aligned} $

$ \begin{aligned} & \Rightarrow x=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4} \\\\ & \text { So, } x \in\left\{0, \frac{\pi}{2}\right\} \\\\ & \therefore x \text { has } 2 \text { elements } \rightarrow P \end{aligned} $

$ \begin{aligned} & \text { (ii) }\left\{x \in\left(\frac{-5 \pi}{18}, \frac{5 \pi}{18}\right), \sqrt{3} \tan 3 x=1\right\} \\\\ & \Rightarrow \tan 3 x=\frac{1}{\sqrt{3}} \\\\ & \Rightarrow x=n \pi+\frac{\pi}{6} \\\\ & \Rightarrow x=\frac{n \pi}{3}+\frac{\pi}{18} \\\\ & \text { So } x \in\left\{\frac{\pi}{18}, \frac{-5 \pi}{18}\right\} \\\\ & \therefore x \text { has 2 elements } \rightarrow P \end{aligned} $

$ \begin{aligned} & \text { (iii) }\left\{x \in\left[\frac{-6 \pi}{5}, \frac{6 \pi}{5}\right], 2 \cos 2 x=\sqrt{3}\right\} \\\\ & \Rightarrow 2 \cos 2 x=\sqrt{3} \\\\ & \Rightarrow \cos 2 x=\frac{\sqrt{3}}{2} \\\\ & \Rightarrow 2 x=2 n \pi \pm \frac{\pi}{6} \\\\ & \Rightarrow x=n \pi \pm \frac{\pi}{12} \\\\ & \text { So } x \in\left\{ \pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12},-\pi \pm \frac{\pi}{12}\right\} \\\\ & \therefore x \text { has } 6 \text { elements } \rightarrow \mathrm{T} \end{aligned} $

$ \begin{aligned} & \text { (iv) }\left\{x \in\left[\frac{-7 \pi}{4}, \frac{7 \pi}{4}\right], \sin x-\cos x=1\right\} \\\\ & \Rightarrow \sin x-\cos x=1 \\\\ & \Rightarrow \sin \left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\\\ & \Rightarrow x-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{4} \\\\ & \Rightarrow x=n \pi+(-1)^n \frac{\pi}{4}+\frac{\pi}{4} \\\\ & \text { So, } \\\\ & x \in\left\{\frac{\pi}{2}, \frac{-3 \pi}{2}, \pi,-\pi\right\} \\\\ & \therefore x \text { has } 4 \text { elements } \rightarrow \mathrm{R} \end{aligned} $

$2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1$

$2\cos x\left( {4\left( {{{\sin }^2}{\pi \over 4} - {{\sin }^2}x} \right) - 1} \right) = 1$

$2\cos x\left( {4\left( {{1 \over 2} - {{\sin }^2}x} \right) - 1} \right) = 1$

$2\cos x(2 - 4{\sin ^2}x - 1) = 1$

$2\cos x(1 - 4{\sin ^2}x) = 1$

$2\cos x(4{\cos ^2}x - 3) = 1$

$4{\cos ^3}x - 3\cos x = {1 \over 2}$

$\cos 3x = {1 \over 2}$

$x \in [0,\pi ]$ $\therefore$ $3x \in [0,3\pi ]$

${(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81$

$ \Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81$

$ \Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}$

taking log of base 32 both side,

$ \Rightarrow $ tan²x = ${\log _{32}}\left( {{{81} \over {33}}} \right)$

$ \Rightarrow $ tan x = $\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)} $

As value of $\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)} $ belongs to (0, 1).

In interval $\,0 \le x \le {\pi \over 4}$ only one solution.

${{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|$

$ \Rightarrow {{{{\cos }^2}x/2 - {{\sin }^2}x/2} \over {(\cos x/2 + \sin x/2)^2}} = \left| {\tan 2x} \right|$

$ \Rightarrow $ ${{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}} = \left| {\tan 2x} \right|$

$ \Rightarrow $ ${{1 - \tan {x \over 2}} \over {1 + \tan {x \over 2}}} = \left| {\tan 2x} \right|$

$ \Rightarrow $ ${{\tan {\pi \over 4} - \tan {x \over 2}} \over {\tan {\pi \over 4} + \tan {x \over 2}}} = \left| {\tan 2x} \right|$

$ \Rightarrow {\tan ^2}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {\tan ^2}2x$

$ \Rightarrow 2x = n\pi \pm \left( {{\pi \over 4} - {\pi \over 2}} \right)$

$ \Rightarrow x = {{ - 3\pi } \over {10}},{{ - \pi } \over 6},{\pi \over {10}}$

or sum $ = {{ - 11\pi } \over 6}$.

$(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$

$ \Rightarrow 2\sin {{5x} \over 2}\left\{ {\cos {{3x} \over 2} + \cos {x \over 2}} \right\} = 0$

$ \Rightarrow 2\sin {{5x} \over 2}\left\{ {2\cos x\cos {x \over 2}} \right\} = 0$

$2\sin {{5x} \over 2} = 0 \Rightarrow {{5x} \over 2} = 0,\pi ,2\pi ,3\pi ,4\pi ,5\pi $

$ \Rightarrow x = 0,{{2\pi } \over 5},{{4\pi } \over 5},{{6\pi } \over 5},{{8\pi } \over 5},2\pi $

$\cos {x \over 2} = 0 \Rightarrow {x \over 2} = {\pi \over 2} \Rightarrow x = \pi $

$\cos x = 0 \Rightarrow x = {\pi \over 2},{{3\pi } \over 2}$

So, sum $ = 6\pi + \pi + 2\pi = 9\pi $

sin⁷x $\le$ sin²x $\le$ 1 ...... (1)

and cos⁷x $\le$ cos²x $\le$ 1 ..... (2)

also sin²x + cos²x = 1

$\Rightarrow$ equality must hold for (1) & (2)

$\Rightarrow$ sin⁷x = sin²x & cos⁷x = cos²x

$\Rightarrow$ sin x = 0 & cos x = 1

or

cos x = 0 & sin x = 1

$\Rightarrow$ x = 0, 2$\pi$, 4$\pi$, ${\pi \over 2},{{5\pi } \over 2}$

$\Rightarrow$ 5 solutions

$x + 2\tan x = {\pi \over 2}$ in [0, 2$\pi$]

$2\tan x = {\pi \over 2} - x$

$2\tan x = {\pi \over 2} - x$

$\tan x = {\pi \over 4} - {x \over 2}$

$y = \tan x$ and $y = {{ - x} \over 2} + {\pi \over 4}$

3 intersection points on the graph.

$ \therefore $ 3 solutions.

${(81)^{{{\sin }^2}x}} + {(81)^{1 - {{\sin }^2}x}} = 30$

${(81)^{{{\sin }^2}x}} + {{81} \over {{{(81)}^{{{\sin }^2}x}}}} = 30$

Let ${(81)^{{{\sin }^2}x}} = t$

$t + {{81} \over t} = 30 \Rightarrow {t^2} + 81 = 30t$

${t^2} - 30t + 81 = 0$

${t^2} - 27t - 3t + 81 = 0$

$(t - 3)(t - 27) = 0$

$t = 3,27$

${(81)^{{{\sin }^2}x}} = 3,{3^3}$

${3^{4{{\sin }^2}x}} = {3^1},{3^3}$

$4{\sin ^2}x = 1,3$

${\sin ^2}x = {1 \over 4},{3 \over 4}$

in [0, $\pi$ ] sin x > 0

$\sin x = {1 \over 2},{{\sqrt 3 } \over 2}$

$x = {\pi \over 6},{{5\pi } \over 6},{\pi \over 3},{{2\pi } \over 3}$

Number of solution = 4

$\sin 2\theta + \tan 2\theta > 0$

$ \Rightarrow \sin 2\theta + {{\sin 2\theta } \over {\cos 2\theta }} > 0$

$ \Rightarrow \sin 2\theta {{(\cos 2\theta + 1)} \over {\cos 2\theta }} > 0 \Rightarrow \tan 2\theta (2{\cos ^2}\theta ) > 0$

Note : $\cos 2\theta \ne 0$

$ \Rightarrow 1 - 2{\sin ^2}\theta \ne \theta \Rightarrow \sin \theta \ne \pm {1 \over {\sqrt 2 }}$

Now, $\tan 2\theta (1 + \cos 2\theta ) > 0$

$ \Rightarrow \tan 2\theta > 0$ (as $\cos 2\theta + 1 > 0$)

$ \Rightarrow 2\theta \in \left( {0,{\pi \over 2}} \right) \cup \left( {\pi ,{{3\pi } \over 2}} \right) \cup \left( {2\pi ,{{5\pi } \over 2}} \right) \cup \left( {3\pi ,{{7\pi } \over 2}} \right)$

$ \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{5\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{7\pi } \over 4}} \right)$

1 - 2sin²x + $\alpha $ sin x = 2$\alpha $ - 7

$ \Rightarrow $ 2sin²x - $\alpha $ sin x + (2$\alpha $ - 8) = 0

$\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}$

$ \Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}$

${{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2$

$ \Rightarrow {{\alpha - 4} \over 2},2$ (rejected)

To exists solutions

$ - 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6$

$ \therefore $ $\alpha \in \left[ {2,6} \right]$

if $\theta $ $ \in $ $\left( {{\pi \over 2},{{2\pi } \over 3}} \right)$ $ \Rightarrow \cos \theta \in \left( { - {1 \over 2},0} \right) \Rightarrow - \cos \theta \in \left( {0,{1 \over 2}} \right)$

$\sin \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)$ and $\cot \theta \in \left( { - {1 \over {\sqrt 3 }},0} \right)$

If $\theta \in \left( {\pi ,{{7\pi } \over 6}} \right) \Rightarrow \cos \theta \in \left( { - 1,{{ - \sqrt 3 } \over 2}} \right) \Rightarrow - \cos \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)$

$\sin \theta \in \left( { - {1 \over 2},0} \right)$ and $\cot \theta \in \left( {\sqrt 3 ,\infty } \right)$

Then in $\theta $ $ \in $ $\left( {{\pi \over 2},{{2\pi } \over 3}} \right)$ $ \Rightarrow $ [sin$\theta $] = 0; [-cos$\theta $] = 0; [cot$\theta $] = -1;

Hence 0 = 0 and -x + y = 0 [have infinitely solutions]

and in $\theta \in \left( {\pi ,{{7\pi } \over 6}} \right)$ $ \Rightarrow $ [sin$\theta $] = -1; [-cos$\theta $] = 0; [cot$\theta $] = 1, 2. 3 ...........;

Then -x - 0.y = 0 $ \Rightarrow $ x = 0

$[\cot \theta ]x + y = 0$

$ \Rightarrow $ 1.x + y = 0 or 2x + y = 0 or ....... each of the line will cut x = 0 at exactly one point.

Hence unique solutions.

$\mathop {\underline {1 + {{\sin }^4}x} }\limits_{ \ge 1} = \mathop {\underline {{{\cos }^2}3x} }\limits_{ \le 1} $

Now for equality to hold sin⁴x = 0 & cos²3x = 1

sin⁴ = 0 $ \Rightarrow $ x = -2$\pi $, - $\pi $, 0, $\pi $, 2$\pi $

All of which satisfy cos²3x = 1 $ \Rightarrow $ 5 solutions

2cos²$\theta $ + 3sin$\theta $ = 0

$ \Rightarrow $ 2(1 - sin²$\theta $) + 3sin$\theta $ = 0

$ \Rightarrow $ 2sin²$\theta $ - 3sin$\theta $ - 2 = 0

$ \Rightarrow $ 2sin²$\theta $ - 4sin$\theta $ + sin$\theta $ - 2 = 0

$ \Rightarrow $ (2sin$\theta $ + 1)(sin$\theta $ - 2) = 0

$ \therefore $ sin$\theta $ = 2 (Not possible)

sin$\theta $ = $ - {1 \over 2}$

$ \therefore $ $\theta $ = n$\pi $ + (-1)ⁿ$\left( { - {\pi \over 6}} \right)$

When n = 0, $\theta $ = ${ - {\pi \over 6}}$

When n = 1, $\theta $ = ${{7\pi } \over 6}$

When n = -1, $\theta $ = ${-{5\pi } \over 6}$

When n = 2, $\theta $ = ${{11\pi } \over 6}$

$ \therefore $ Sum of all solutions in [–2$\pi $, 2$\pi $] is

= $ - {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6}$ = 2$\pi $

sin²2$\theta $ + cos⁴2$\theta $ = ${3 \over 4}, $$\theta $ $ \in $ $\left( {0,{\pi \over 2}} \right)$

$ \Rightarrow $  1 $-$ cos²2$\theta $ + cos⁴2$\theta $ = ${3 \over 4}$

$ \Rightarrow $  4cos2$\theta $ $-$ 4cos²2$\theta $ + 1 = 0

$ \Rightarrow $  (2cos²2$\theta $ $-$ 1)² = 0

$ \Rightarrow $  cos²2$\theta $ = ${1 \over 2}$ = cos²${{\pi \over 4}}$

$ \Rightarrow $  2$\theta $ = n$\pi $ $ \pm $ ${\pi \over 4}$, n $ \in $ I

$ \Rightarrow $  $\theta $ = ${{n\pi } \over 2} \pm {\pi \over 8}$

$ \Rightarrow $  $\theta $ = ${\pi \over 8},{\pi \over 2} - {\pi \over 8}$

Sum of solutions ${\pi \over 2}$

sin x $-$ sin 2x + sin 3x = 0        $x \in \left[ {0,{\pi \over 2}} \right)$

$ \Rightarrow $  (sin3x + sinx) $-$ sin2x = 0

$ \Rightarrow $  2sin2x.cos2x $-$ sin2x = 0

$ \Rightarrow $  sin2x (2cosx $-$ 1) = 0

sin 2x = 0

x = 0

and cos x = ${1 \over 2}$

and x = ${\pi \over 3}$

two solutions

For Z = {x : g(x) = 0}, x > 0

$ \because $ g(x) = cos(2$\pi $ sin x) = 0

$ \Rightarrow $ $2\pi \sin x = (2n + 1){\pi \over 2},\,n \in $ Integer

$ \Rightarrow $ $\sin x = - {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$ [$ \because $ sin x $ \in $ [$-$1, 1]]

here values of sin x, $ - {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$ are in an A.P. but corresponding values of x are not in an AP so, (iii) $ \to $ R.

For W = {x : g'(x) = 0}, x > 0

So, g'(x) = $-$2 $\pi $ cos x sin(2$\pi $ sin x) = 0

$ \Rightarrow $ either cos x = 0 or sin(2$\pi $ sin x) = 0

$ \Rightarrow $ either $x = (2n + 1){\pi \over 2}$ or 2$\pi $ sin x = n$\pi $, n$ \in $ Integers.

$ \because $ $2\pi \sin x = nx$

$ \Rightarrow $ $\sin x = {n \over 2} = - 1, - {1 \over 2},0,{1 \over 2},1$ {$ \because $ sin x $ \in $[$-$1, 1)}

$ \because $ $x = n\pi ,\,(2n + 1){\pi \over 2}$ or $x = n\pi + {( - 1)^n}\left( { \pm {\pi \over 6}} \right)$

$ \Rightarrow $ (iv) $ \to $ P, R, S

Hence, option (a) is correct.

For, X = {x : f(x) = 0}, x > 0

Now, f(x) = 0

$ \Rightarrow $ sin($\pi $ cos x) = 0, x > 0

$ \Rightarrow $ $\pi $ cos x = n$\pi $, n $ \in $ Integer.

$ \Rightarrow $ cos x = n

$ \Rightarrow $ cos x = $-$1, 0, 1

{$ \because $ cos x $ \in $[$-$1, 1]}

When cos x = ± 1$ \Rightarrow $ x = n$\pi $

When cos x = 0 $ \Rightarrow $ x = (2n + 1)${{\pi \over 2}}$

Hence, (i) $ \to $ (P), (Q)

For, Y = {x : f'(x) = 0}, x > 0

Now, f'(x) = 0

$ \Rightarrow $ $-$$\pi $ sin x cos($\pi $ cos x) = 0

$ \Rightarrow $ either sin x = 0 $ \Rightarrow $ x = n$\pi $, n is an integer,

or cos($\pi $ cos x) = 0

$ \Rightarrow $ $\pi $ cos x = (2n + 1)${{\pi \over 2}}$, n is an integer

$ \Rightarrow $ cos x = ${{{2n + 1} \over 2}}$

$ \Rightarrow $ $\cos x = \pm {1 \over 2}$ {$ \because $ cos x $ \in $[$-$1, 1]}

$ \Rightarrow $ x = $2n\pi \pm {\pi \over 3}$ or $2n\pi \pm {{2\pi } \over 3}$, n is an integer.

So, (ii) $ \to $ (Q), (T)

Hence, option (a) is correct.

As we know,

$\cos \left( {x + y} \right)\cos \left( {x - y} \right) = {\cos ^2}x - {\sin ^2}y$

$\therefore$ ${\mathop{\rm cos}\nolimits} \left( {{\pi \over 6} + x} \right)cos\left( {{\pi \over 6} - x} \right) = {\cos ^2}\left( {{\pi \over 6}} \right) - {\sin ^2}x$

Given, $8\cos x\left( {\cos \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right) - {1 \over 2}} \right) = 1$

$ \Rightarrow 8\cos x\left( {{{\cos }^2}{\pi \over 6} - {{\sin }^2}x - {1 \over 2}} \right) = 1$

$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - {{\sin }^2}x} \right) = 1$

$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - 1 + {{\cos }^2}x} \right) = 1$

$ \Rightarrow 8\cos x\left( {{{\cos }^2}x - {3 \over 4}} \right) = 1$

$ \Rightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1$

(Taking $4\cos x$ inside the bracket).

We know,

$4{\cos ^3}x - 3\cos x = \cos 3x$

$ \Rightarrow 2\cos 3x = 1$

$ \Rightarrow \cos 3x = {1 \over 2}$

$\therefore$ $\,\,\,$ $3x = 2n\pi \pm {\pi \over 3}$

So, $x = {{2n\pi } \over 3} \pm {\pi \over 9}$

At $n=0,$ $x = + {\pi \over 9}$ as $x \in \left[ {0,\pi } \right]$

At $n=1,$ $x = {{2\pi } \over 3} + {\pi \over 9}$

$\therefore$ $\,\,\,$ $x = {{2\pi } \over 3} + {\pi \over 9}$

and $x = {{2\pi } \over 3} - {\pi \over 9}$

At $n = 2,$ $x = {{4\pi } \over 3} \pm {\pi \over 9}$

Which does not belongs to $\left[ {0,\pi } \right]$

So, possible values of $x$ is $ = {\pi \over 9},{{2\pi } \over 3} + {\pi \over 9},{{2\pi } \over 3} - {\pi \over 3}$

Sum of the solutions $ = {\pi \over 9} + {{2\pi } \over 3} + {\pi \over 9} + {{2\pi } \over 3} - {\pi \over 9}$ v

$ = {{4\pi } \over 3} + {\pi \over 9}$ $ = {{13\,\pi } \over 9}$

According to the question,

$k\pi = {{13\pi } \over 9}$

$\therefore\,\,\,$ $k = {{13} \over 9}$

sin 3x = cos 2x

$ \Rightarrow $$\,\,\,$ 3 sin x $-$ 4 sin³ x = 1 $-$ 2 sin² x

$ \Rightarrow $$\,\,\,$ 4 sin³ x $-$ 2 sin² x $-$ 3 sin x + 1 = 0

$ \Rightarrow $$\,\,\,$ sin x = 1, ${{ - 2 \pm 2\sqrt 5 } \over 8}$

In the interval $\left( {{\pi \over 2},\pi } \right)$, sin x = ${{ - 2 + 2\sqrt 5 } \over 8}$

So, there is only one solution.

cos(P + Q) + cos(Q + R) + cos(R + P)

= $-$ (cosR + cosP + cosQ)

Max. of cosP + cosQ + cosR = ${3 \over 2}$

Min. of cos(P + Q) + cos(Q + R) + cos(R + P) is = $ - {3 \over 2}$

We have,

$ \left|\sqrt{2 \sin ^4 x+18 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right|=1 $

That is,

$ f(x)=\left|\sqrt{2 \sin ^4 x+8 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right| $

Now, $f(x)=f\left(\frac{\pi}{2}-x\right) $

$\Rightarrow f(x)$ is symmetric about $x=\frac{\pi}{4}$.

If $f(x)$ has solution in $(0, \pi / 4)$, then in $(0,2 \pi)$, there are eight solutions.

$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$

$ \Rightarrow $ $(\cos x + \cos 3x)$ + $(\cos 2x + \cos 4x)$ = 0

$ \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0$

$ \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0$

$\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0$

$x = \pi ,{\pi \over 2},{{3\pi } \over 2},{\pi \over 5},{{3\pi } \over 5},{{7\pi } \over 5},{{9\pi } \over 5}$

It is given that,

$\sum\limits_{k = 1}^{13} {{1 \over {\sin \left( {{\pi \over 4} + {{(k - 1)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}} $

Let $\alpha = {\pi \over 4}$ and $\beta = {\pi \over 6}$. Therefore,

$\sum\limits_{k = 1}^{13} {{1 \over {\sin (\alpha + k\beta )sin(\alpha + (k - 1)\beta )}}} $

$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {{{\sin ((\alpha + k\beta ) - (\alpha + (k - 1)\beta ))} \over {\sin (\alpha + k\beta )\sin (\alpha + (k - 1)\beta )}}} $

$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {(\cot (\alpha + (k - 1)\beta ) - \cot (\alpha + k\beta ))} $

$ = {1 \over {\sin \beta }}\{ [\cot (\alpha ) - \cot (\alpha + \beta )] + [\cot (\alpha + \beta ) - \cot (\alpha + 2\beta )] + ...... + [\cot (\alpha + 12\beta ) - \cot (\alpha + 13\beta )]\} $

$ = {1 \over {\sin \beta }}(\cot \alpha - \cot (\alpha + 13\beta ))$

$ = {1 \over {\sin (\pi /6)}}\left( {\cot {\pi \over 4} - \cot \left( {{\pi \over 4} + {{13\pi } \over 6}} \right)} \right)$

$ = 2(1 - 2 + \sqrt 3 ) = 2(\sqrt 3 - 1)$