Trigonometric Equations

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

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

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

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

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

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

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

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

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

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

$ \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.

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.

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.

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

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)$

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.

$2{\sin ^2}x + 5\sin x - 3 = 0$

$ \Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0$

$\sin x = {1 \over 2}$ and $\,\,\sin x \ne - 3$

Given that $x \in \left[ {0,3\pi } \right]$

So possible values of x are $30^\circ $, $150^\circ $, $390^\circ $, $510^\circ $. That means x have 4 values.

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

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

$t_{1} < t_{4}$

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

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