Trigonometric Equations

(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 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}} $

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.

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.

It is given, that for non-negative integers 'n',

$f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }} $

$= {{\sum\limits_{k = 0}^n {\left( {\cos {\pi \over {n + 2}} - \cos \left( {{{2k + 3} \over {n + 2}}\pi } \right)} \right)} } \over {\sum\limits_{k = 0}^n {\left( {1 - \cos \left( {{{2k + 2} \over {n + 2}}\pi } \right)} \right)} }}$

[$ \because $ $2\sin A\sin B = \cos (A - B) - \cos (A + B)\,and\,2si{n^2}A = 1 - \cos 2A$]

$={{\left( {\cos \left( {{\pi \over {n + 2}}} \right)} \right)\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{3\pi } \over {n + 2}} + \cos {{5\pi } \over {n + 2}} + \cos {{7\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 3} \over {n + 2}}\pi } \right)} \right\}} } \over {\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{2\pi } \over {n + 2}} + \cos {{4\pi } \over {n + 2}} + \cos {{6\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 2} \over {n + 2}}\pi } \right)} \right\}} }}$

$={{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)} \over {(n + 1) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)}}$

[$ \because $ $\cos (\alpha ) + \cos (\alpha + \beta ) + cos(\alpha + 2\beta ) + ... + cos(\alpha + (n - 1)\beta ) = {{\sin \left( {{{n\beta } \over 2}} \right)} \over {\sin \left( {{\beta \over 2}} \right)}}\cos \left. {\left( {{{2\alpha + (n - 1)\beta } \over 2}} \right)} \right]$

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

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

$ = {{(n + 2)\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 2)}} = \cos \left( {{\pi \over {n + 2}}} \right)$

$ \Rightarrow f(n) = \cos \left( {{\pi \over {n + 2}}} \right)$

Now, $f(6) = \cos \left( {{\pi \over 8}} \right)$

$ \because $ $\alpha = \tan ({\cos ^{ - 1}}f((6))) = \tan \left( {{\pi \over 8}} \right)$

$\left\{ \matrix{ {\cos ^{ - 1}}\cos x = x \hfill \cr if\,x \in \left( {0,\,{\pi \over 2}} \right) \hfill \cr} \right\}$

$ = \sqrt 2 - 1$

$ \Rightarrow (\alpha + 1) = \sqrt 2 \Rightarrow {(\alpha + 1)^2} = 2 \Rightarrow {\alpha ^2} + 2\alpha + 1 = 2$

$ \Rightarrow {\alpha ^2} + 2\alpha - 1 = 0$

Now, $f(4) = \cos \left( {{\pi \over {4 + 2}}} \right) = \cos \left( {{\pi \over 6}} \right) = {{\sqrt 3 } \over 2}$,

Now, $\sin (7{\cos ^{ - 1}}f(5)) = \sin \left( {7{{\cos }^{ - 1}}\left( {\cos \left( {{\pi \over {5 + 2}}} \right)} \right)} \right) = \sin \left( {7\left( {{\pi \over 7}} \right)} \right) = \sin \pi = 0$

and Now, $\mathop {\lim }\limits_{n \to \infty } f(x) = \mathop {\lim }\limits_{n \to \infty } \cos {\pi \over {n + 2}} = \cos 0 = 1$

Hence, options (a), (b) and (c) are correct.

We have,

In $\Delta $PQR

$\angle PQR = 30^\circ $

$PQ = 10\sqrt 3 $

$QR = 10$



By cosine rule

$\cos 30^\circ = {{P{Q^2} + Q{R^2} - P{R^2}} \over {2PQ.QR}}$

$ \Rightarrow {{\sqrt 3 } \over 2} = {{300 + 100 - P{R^2}} \over {200\sqrt 3 }}$

$ \Rightarrow 300 = 300 + 100 - P{R^2}$

$ \Rightarrow PR = 10$

Since, $PR = QR = 10$

$ \therefore $ $\angle QPR = 30^\circ $ and $\angle QRP = 120^\circ $

Area of $\Delta PQR = {1 \over 2}PQ.QR.\sin 30^\circ $

$ = {1 \over 2} \times 10\sqrt 3 \times 10 \times {1 \over 2} = 25\sqrt 3 $

Radius of incircle of

$\Delta PQR = {{Area\,of\,\Delta PQR} \over {Semi - perimetre\,of\,\Delta PQR}}$

i.e. $r = {\Delta \over s} = {{25\sqrt 3 } \over {{{10\sqrt 3 + 10 + 10} \over 2}}} = {{25\sqrt 3 } \over {5(\sqrt 3 + 2)}}$

$ \Rightarrow r = 5\sqrt 3 (2 - \sqrt 3 ) = 10\sqrt 3 - 15$

and radius of circumcircle

$(R) = {{abc} \over {4\Delta }} = {{10\sqrt 3 \times 10 \times 10} \over {4 \times 25\sqrt 3 }} = 10$

$ \therefore $ Area of circumcircle of

$\Delta PQR = \pi {R^2} = 100\pi $

Hence, option (b), (c) and (d) are correct answer.

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}$

$2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$

or $4(\cos \beta - \cos \alpha ) + 2\cos \alpha \cos \beta = 2$

$ \Rightarrow 1 - \cos \alpha + \cos \beta - \cos \alpha \cos \beta $

$ = 3 + 3\cos \alpha - 3\cos \beta - 3\cos \alpha \cos \beta $

$ \Rightarrow (1 - \cos \alpha )(1 + \cos \beta )$

$ = 3(1 + \cos \alpha )(1 - \cos \beta )$

$ \Rightarrow {{(1 - \cos \alpha )} \over {(1 + \cos \alpha )}} = {{3(1 - \cos \beta )} \over {1 + \cos \beta }}$

$ \Rightarrow {\tan ^2}{\alpha \over 2} = 3{\tan ^2}{\beta \over 2}$

$ \therefore $ $\tan {\alpha \over 2} \pm \sqrt 3 \tan {\beta \over 2} = 0$

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)$

Let us consider

$ S=\left\{x \in(-\pi, \pi), x \neq 0, \pm \frac{\pi}{2}\right\} $

The given equation is

$ \begin{aligned} & \sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0 \\\\ & \Rightarrow \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x=2 \cos 2 x \\\\ & \Rightarrow \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x \\\\ & \Rightarrow \cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x \\\\ & \Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right) \\\\ & \Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right) \quad\quad(x \in I) \end{aligned} $

Case 1: When

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

we have $x=2 n \pi-\frac{\pi}{3}$.

If $n=0$, we get $x=-\frac{\pi}{3}$.

If $n=1$, we get $x=2 \pi-\frac{\pi}{3}$.

If $n=-1, x=-2 \pi-\frac{\pi}{3}$.

Case 2: When

$2 x=2 n \pi-x+\frac{\pi}{3}$, we get $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$.

If $n=0$, we get $x=\frac{\pi}{9}$.

If $n=1$, we get $x=\frac{2 \pi}{3}+\frac{\pi}{9}$.

If $n=2$, we get $n=2 x=\frac{4 \pi}{3}+\frac{\pi}{9}$.

If $n=-1$, we get $x=\frac{-2 \pi}{3}+\frac{\pi}{9}$.

Therefore, the sum of all distinct solutions of the given equation is

$ \frac{-\pi}{3}+\frac{\pi}{9}+\frac{2 \pi}{3}+\frac{\pi}{9}-\frac{2 \pi}{3}+\frac{\pi}{9}=0 $

$\begin{aligned} & \text { Given, } \sin x+2 \sin 2 x-\sin 3 x=3 \quad \text{... (i)}\\ & \Rightarrow \quad-[\sin 3 x-\sin x]+2 \sin 2 x=3 \end{aligned}$

We know

$\begin{aligned} & \sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cdot \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) \\ & \text { and } \sin 2 \theta=2 \sin \theta \cdot \cos \theta \\ & \therefore-2 \cos \left(\frac{3 x+x}{2}\right) \cdot \sin \left(\frac{3 x-x}{2}\right)+2 \cdot 2 \sin x \cdot \cos x=3 \\ & \Rightarrow \quad-2 \cos 2 x \cdot \sin x+4 \sin x \cdot \cos x=3 \\ & \Rightarrow \quad-2 \sin x[\cos 2 x-2 \cos x]=3 \\ & \Rightarrow \quad-2 \sin x\left[2 \cos ^2 x-1-2 \cos x\right]=3 \\ & \Rightarrow \quad-4 \sin x\left[\cos ^2 x-2 \cdot \frac{1}{2} \cos x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-\frac{1}{2}\right]=3 \end{aligned}$

$\begin{array}{lr} \Rightarrow & -4 \sin x\left[\left(\cos x-\frac{1}{2}\right)^2-\frac{3}{4}\right]=3 \\ \Rightarrow & -4 \sin x\left(\cos x-\frac{1}{2}\right)^2+3 \sin x=3 \\ \Rightarrow & 3 \sin x-3=4 \sin x\left(\cos x-\frac{1}{2}\right)^2 \end{array}$

Draw the graph of $y=3 \sin x-3$ and $y=\left(\cos x-\frac{1}{2}\right)^2$ in the interval $x \in(0, \pi)$

It is clear that the graphs $y=3 \sin x-3$ and $y=\left(\cos x-\frac{1}{2}\right)$ does not intersect in $x \in(0, \pi)$ Hence, the given equation $\sin x+2 \sin 2 x-\sin 3 x=3$ does not have any solution in the interval $x \in(0, \pi)$

Hint:

(i) Recall the formula $\sin 2 x=2 \sin x \cdot \cos x, \cos 2 x=2 \cos ^2 x-1$ and $\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cdot \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$

(ii) The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$ in the given interval.

$\text { We know the graph of } y=x \sin x \text { is }$

And the graph of $y=\cos x$ is

Now add the graph of $y=x \sin x$ and $y=\cos x$ and also draw the graph of $y=x^2$

It is clear that the graph of $y=x^2$ and $y=x \sin x +\cos x$ intersect at two points

$\therefore$ The equation $x^2=x \sin x+x$ satisfy for two values of $x$.

Hints:

(i) Recall the graph of $y=x \sin x, y=\cos x$ and $y=x^2$.

(ii) Recall the method of addition of two graphs.

(iii) The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$

Given, ,$f(x)=x \cdot \sin \pi x, x>0$

$\Rightarrow \quad f^{\prime}(x)=1 \cdot \sin \pi x+x \cdot \pi \cos \pi x$

Apply $f^{\prime}(x)=0$

$\begin{aligned} & \Rightarrow \sin \pi x+\pi x \cos \pi x=0 \\ & \Rightarrow \sin \pi x=-\pi x \cos \pi x \\ & \Rightarrow \frac{\sin \pi x}{\cos \pi x}=-\pi x \\ & \Rightarrow \tan \pi x=-\pi x \end{aligned}$

Draw the graph of $y=\tan \pi x$ and $y=-\pi x$ for $x>0$

It is clear that $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point if $\frac{1}{2} < x < 1$ as $\frac{3}{2} < x < 2$ as $\frac{5}{2} < x < 3$ or.....

So, $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point is $x \in\left(n+\frac{1}{2}, n+1\right)$ as $(n, n+1)$.

Hints:

The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$

Given, $\tan (2 \pi-\theta)>0$

$\begin{aligned} & \Rightarrow 0<(2 \pi-\theta)<\frac{\pi}{2} \text { or } \pi<(2 \pi-\theta)<\frac{3 \pi}{2} \\ & \Rightarrow \frac{3 \pi}{2}<\theta<2 \pi \text { or } \frac{\pi}{2}<\theta<\pi \quad \text{... (i)} \end{aligned}$

Also, $-1<\sin \theta<\frac{-\sqrt{3}}{2}$

$\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3} \quad \text{... (ii)}$

From (i) and (ii),

$\Rightarrow \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\quad \text{... (iii)}$

Now,

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

$\begin{aligned} & \Rightarrow \quad 2 \cos \theta(1-\sin \phi) \\ &=\sin \theta \cdot \sin \theta\left(\frac{1}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad 2 \cos (1-\sin \phi)=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cdot \cos \phi \cdot \sin \theta}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}-1 \\ & \Rightarrow \quad 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1 \\ & \Rightarrow \quad 2 \cos \theta+1=2 \sin (\theta+\phi) \\ & \text { As } \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) \quad \text{[from (iii)]}\\ & \Rightarrow 2 \cos \theta+1 \in(1,2) \\ & \Rightarrow 1<2 \sin (\theta+\phi)<2 \\ & \Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1 \end{aligned}$

$\begin{aligned} & \text { Now, } \theta, \phi \in[0,2 \pi] \\ & \theta+\phi \in[0,4 \pi] \\ & \Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) \text { or } \theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) \\ & \Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta \text { or } \frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta \\ & \Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) \\ & {\left[\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right] } \end{aligned}$

$P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} $

$ \Rightarrow \cos \theta \left( {\sqrt 2 + 1} \right) = \sin \theta $

$ \Rightarrow \tan \theta = \sqrt 2 + 1$ ..... (i)

$Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} $

$ \Rightarrow \sin \theta \left( {\sqrt 2 - 1} \right) = \cos \theta $

$ \Rightarrow \tan \theta = {1 \over {\sqrt 2 - 1}} \times {{\sqrt 2 + 1} \over {\sqrt 2 + 1}}$

$ = \left( {\sqrt 2 + 1} \right)$ ...... (ii)

$\therefore$ $P = Q$

(a) We have

$2{\sin ^2}\theta + 4{\sin ^2}\theta {\cos ^2}\theta = 2$

${\sin ^2}\theta + 2{\sin ^2}\theta (1 - {\sin ^2}\theta ) = 1$

$3{\sin ^2}\theta - 2{\sin ^4}\theta - 1 = 0$

$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 2 }}, \pm 1$

$ \Rightarrow \theta = {\pi \over 4},{\pi \over 2}$

(B) Let $y = {{3x} \over \pi } \Rightarrow {1 \over 2} \le y \le 3\forall x \in \left[ {{\pi \over 4},\pi } \right]$

Now, $f(y) = [2y]\cos [y]$.

The critical points are

$y = {1 \over 2},y = 1,y = {3 \over 2}$ and $y = 3$

$\Rightarrow$ points of discontinuity $\left\{ {{\pi \over 6},{\pi \over 3},{\pi \over 2},\pi } \right\}$.

(C) $\left| {\matrix{ 1 & 1 & 0 \cr 1 & 2 & 0 \cr 1 & 1 & \pi \cr } } \right| = \pi \Rightarrow $ volume of parallelepiped = $\pi$.

(D) We have

$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3 $

$ \Rightarrow \sqrt {2 + 2\cos \alpha } = \sqrt 3 $

$ \Rightarrow 2 + 2\cos \alpha = 3$

$ \Rightarrow \alpha = {\pi \over 3}$

Given solutions

${1 \over {\sin (\pi /4)}}\left[ {{{\sin (\theta + \pi /4 - \theta )} \over {\sin \theta \,.\,\sin (\theta + \pi /4)}} + {{\sin (\theta + \pi /2 - (\theta + \pi /4))} \over {\sin (\theta + \pi /4)\,.\,(\theta + \pi /2)}}\, + \,...\, + \,{{\sin ((\theta + 3\pi /2) - (\theta + 5\pi /4))} \over {\sin (\theta + 3\pi /2)\,.\,\sin (\theta + 5\pi /4)}}} \right] = 4\sqrt 2 $

$ \Rightarrow \sqrt 2 [\cos \theta - \cot (\theta + \pi /4) + \cot (\theta + \pi /4) - \cot (\theta + \pi /2)\, + \,...\, + \,\cot (\theta + 5\pi /4) - \cot (\theta + 3\pi /2)] = 4\sqrt 2 $

$ \Rightarrow \tan \theta + \cot \theta = 4 \Rightarrow \tan \theta = 2 \pm \sqrt 3 $

$ \Rightarrow \theta = {\pi \over {12}}$ or ${{5\pi } \over {12}}$

It is given that

${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5}$

$3{\sin ^4}x + 2{(1 - {\sin ^2}x)^2} = {6 \over 5}$

$ \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0$

$ \Rightarrow {\sin ^2}x = {2 \over 5}$ and ${\cos ^2}x = {3 \over 5}$

Hence, ${\tan ^2}x = {2 \over 3}$

Therefore, ${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$

(A) Let $y = {{{x^2} + 2x + 4} \over {x + 2}}$

${{dy} \over {dx}} = {{{x^2} + 4x} \over {{{(x + 2)}^2}}} = 0$

$x = 0, - 4$

${{{d^2}y} \over {d{x^2}}} = {8 \over {{{(x + 2)}^3}}}$

At $x = 0,{{{d^2}y} \over {d{x^2}}}$ is true

$\therefore$ y is min. when $x = 0$

$\therefore$ ${y_{\min }} = 2$

(A) - (iii)

(B) As A is symmetric and B is skew symmetric matrix

We should have

A$^+$ = A and B$^+$ = $-$B ...... (i)

Also given that

(A + B) (A $-$ B) = (A $-$ B) (A + B)

A$^2$ $-$ AB + BA $-$ B$^2$ = A$^2$ + AB $-$ AB $-$ B$^2$

2BA = 2AB or AB = BA ...... (ii)

Now, given that

(AB)$^t$ = ($-$1)$^k$AB

(BA)$^t$ = ($-$1)$^k$AB (Using equation (i))

$\Rightarrow$ K should be an odd no.

$\therefore$ B - (ii, iv)

(C) Given that,

$a = {\log _3}{\log _3}2$

$ \Rightarrow {\log _3}2 = {3^a} \Rightarrow {{{1_x}} \over {{{\log }_2}^3}} = {3^a}$

Or ${\log _2}^3 = {3^{ - a}}$

$3 = {2^{(3 - a)}}$

Now, $ < {2^{( - k + 3 - a)}} < 2 \Rightarrow 1 < {2^{ - 2}}.\,{2^{3 - a}} < 2$

$ \Rightarrow 1 < {2^{ - k}}\,.\,3 < 2$ (using eq. (i))

$ = {1 \over 3}\,.\, < {2^{ - k}} < {2 \over 3} \Rightarrow {3 \over 2} < {2^k} < 3$

$ \Rightarrow k = 1$

$\therefore$ k is less than 2 and 3.

$\therefore$ (C) - (iii, iv)

(D) Given that,

$\sin \theta = \cos \phi $

$\cos \left( {{\pi \over 2} - \theta } \right) = \cos \phi $

$ = {\pi \over 2} - \theta = 2n\pi \pm \phi ,n \in Z$

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

$ = {1 \over \pi }\left( {\theta \pm \phi - {\pi \over 2}} \right) = - 2n$

$\therefore$ Here, possible value of ${1 \over \pi }\left( {\theta \pm \phi - {\pi \over 2}} \right)$ are 0 and 2 for $n = 0, - 1$

$\therefore$ (D) - p, r

Given,

$2{\sin ^2}\theta - \cos 2\theta = 0$ and

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

$2{\sin ^2}\theta - \cos 2\theta = 0$ ..... (i)

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

$4{\sin ^2}\theta - 1 = 0$

$4{\sin ^2}\theta = 1$

${\sin ^2}\theta = {1 \over 4}$

$\sin \theta = \sqrt {{1 \over 4}} \Rightarrow \sin \theta = \pm {1 \over 2}$

And, $2{\cos ^2}\theta - 3\sin \theta = 0$ ..... (ii)

$2(1 - {\sin ^2}\theta ) - 3\sin \theta = 0$

$2 - 2{\sin ^2}\theta - 3\sin \theta = 0$

$2{\sin ^2}\theta + 3\sin \theta - 2 = 0$

$\sin \theta = {{ - 3 \pm \sqrt {{3^2} - 4 \times 2 \times ( - 2)} } \over {2 \times 2}}$

$\sin \theta = {{ - 3 \pm \sqrt {9 + 16} } \over 4}$ (using quadratic formula)

$\sin \theta = {{ - 3 \pm 5} \over 4}$ or $\sin \theta = {{ - 3 + 5} \over 4},{{ - 3 - 5} \over 4}$

$\sin \theta = {1 \over 2}$ or $\sin \theta = - 2$ ($ - 1 \le \sin \theta \le 1$)

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

So, the solution of the pair of equation is

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

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

$\sin \theta = \sin \left( {\pi - {\pi \over 6}} \right)$ [$\because$ $\sin (\pi - x) = \sin x$]

$\theta = {\pi \over 6}$ or ${{5\pi } \over 6}$

$\theta = {\pi \over 6},{{5\pi } \over 6},\theta \in [0,2\pi ]$

So, the number of solutions are 2.

$\begin{aligned} & 0 < \theta < \frac{\pi}{4} \\ & \therefore \quad 0 < \tan \theta < 1 \\ & \text { Whereas } 1 < \cot \theta < \infty \\ & t_{1}=(\tan \theta)^{\tan \theta} \\ & \log t_{1}=\tan \theta \log (\tan \theta) \\ & =\tan \theta \log \left(\frac{1}{\cot \theta}\right) \\ & =-\tan \theta \log (\cot \theta) \\ & \Rightarrow \quad t_{1}=-\cot \theta^{\tan \theta} \\ & =-t_{3} \\ & \Rightarrow \quad t_{1} < t_{3} \\ & t_{2}=(\tan \theta)^{\cot \theta} \\ & \log t_{2}=\cot \theta \log (\tan \theta) \\ & =-\cot \theta \log (\cot \theta) \\ & =-\log (\cot \theta)^{\cot \theta} \\ & \therefore \quad t_{2}=-t_{4} \\ & \Rightarrow \quad t_{2} < t_{4} \end{aligned}$

$\text { And also }(\tan \theta)^{\tan \theta} < 1 \text { and }(\cot \theta)^{\cot \theta} > 1$

$t_{1} < t_{4}$

$\text { hence, } \quad t_{4} > t_{3}> t_{1} > t_{2}$

Shortcut Method:

$\begin{aligned} & \tan \theta < 1 \text { and } \cot \theta > 1 \\ & \Rightarrow \quad(\tan \theta)^{\tan \theta}<1 \text { and }(\cot \theta)^{\cot \theta} > 1 \\ & t_{4} > t_{1} \\ & \text { Only (B) have option containing } t_{4} > t_{1} \end{aligned}$

We have the inequality: $2 \sin^2 \theta - 5 \sin \theta + 2 > 0$

Step 1: Let $\sin \theta = t$

The inequality becomes: $2t^2 - 5t + 2 > 0$

Step 2: Factor the Quadratic

We can write: $2t^2 - 5t + 2 = (t-2)(2t-1)$

So, $(t-2)(2t-1) > 0$

Step 3: Find When the Product is Positive

A product $(t-a)(t-b) > 0$ when $t < \min(a,b)$ or $t > \max(a,b)$.

Here, the roots are $t=2$ and $t=\frac{1}{2}$, so: $t < \frac{1}{2} \quad \text{or} \quad t > 2$

Step 4: Relate Back to $\sin \theta$

So, $\sin \theta < \frac{1}{2} \quad \text{or} \quad \sin \theta > 2$

But $\sin \theta$ cannot be bigger than $1$ for any real $\theta$, so $\sin \theta > 2$ is not possible.

So the only valid option is: $\sin \theta < \frac{1}{2}$

Step 5: Find the Values of $\theta$

For $0 < \theta < 2\pi$, $\sin \theta = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.

So, $\sin \theta < \frac{1}{2}$ when $\theta$ is between $0$ and $\frac{\pi}{6}$, or between $\frac{5\pi}{6}$ and $2\pi$.

So the solution is: $\theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right)$