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.

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

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} $

Applying sine rule in $\triangle P Q R$,

$ \begin{aligned} & \frac{\alpha}{\sin 30^{\circ}}=\frac{1}{\sin 80^{\circ}} \\\\ \Rightarrow & \alpha=\frac{1}{2 \sin 80^{\circ}} \end{aligned} $

Applying sine rule in $\triangle P R S$,

$ \begin{aligned} & \frac{\beta}{\sin 40^{\circ}}=\frac{1}{\sin \theta} \\\\ \Rightarrow \beta \sin \theta=& \sin 40^{\circ} \end{aligned} $

From (i) and (ii),

$ 4 \alpha \beta \sin \theta=4 \cdot \frac{1}{2 \sin 80^{\circ}} \cdot \sin 40^{\circ}=\frac{1}{\cos 40^{\circ}} $

Using $\cos 0^{\circ}=1, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$ and $\cos 60^{\circ}=\frac{1}{2}$

Only Options (A) and (B) are correct.

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

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

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

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

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

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

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

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

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

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

$ \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 $