and period of $|\sin 2x| = \frac{\pi }{2}$.
Hence period $= \frac{{2\pi }}{2} = \pi $.
Hence period $ = \frac{{2\pi }}{3}$.
hence period $= 2\pi $.
$= \sin \left( {\frac{1}{n}(2n\pi + x)} \right)$
==> Period of the function $\sin \left( {\frac{x}{n}} \right)$ is $2n\pi $.
==> $2n\pi = 4\pi \Rightarrow n = 2$.
The period of the function in option $(b)$ is $24.$
The period of the function in option $(c)$ is $2\pi $.
Period of $\cos \,\left( {\frac{{\pi x}}{2}} \right) = \frac{{2\pi }}{{\pi /2}} = 4$
$\therefore $ Period of $\sin \frac{{\pi x}}{2} + \cos \frac{{\pi x}}{2} = $$L.C.M. of (4, 4)=4.$
$\tan 6\theta = \frac{{\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta }}{{1 - \sum \tan \theta \tan 2\theta }}$
$= 0, $ (from the given condition)
$ \Rightarrow $ $6\theta = n\pi \Rightarrow \theta = n\pi /6$.
Trick : In such type of problems, the general value of $\theta $ is given by $\frac{{n\pi }}{{{\rm{sum \,of\, number\, of\, }}\theta }}$.
So the general value of $\theta $ is $\frac{{n\pi }}{{1 + 2 + 3}} = \frac{{n\pi }}{6}$.
$ \Rightarrow $ $\cos \theta = \frac{{(\sqrt 2 + 1) \pm \sqrt {{{(\sqrt 2 + 1)}^2} - \frac{8}{{\sqrt 2 }}} }}{4}$
$ \Rightarrow $ $\cos \theta = \cos \left( {\frac{\pi }{4}} \right)$
$ \Rightarrow $ $\theta = 2n\pi \pm \frac{\pi }{4}$.
Trick : Since $\theta = \frac{\pi }{4}$ satisfies the equation and therefore the general value should be $2n\pi \pm \frac{\pi }{4}$.
==> $3(\sin \theta - \cos \theta ) = 2\sin 2\theta $
Squaring both sides, we get $9(1 - S) = 4{S^2},$
where $S = \sin 2\theta $ or $4{S^2} + 9S - 9 = 0$.
$\therefore $ $\,(S + 3)\,(4S - 3) = 0$ or $S = \frac{3}{4}$ as $S \ne - 3$
or $\sin 2\theta = \frac{3}{4} = \sin \alpha $
$\therefore $ $2\theta = n\pi + {( - 1)^n}\alpha $
or $\theta = \frac{1}{2}\,\left[ {n\pi + {{( - 1)}^n}{{\sin }^{ - 1}}\left( {\frac{3}{4}} \right)} \right]$.
Then $r = \sqrt {{{(\sqrt 3 + 1)}^2} + {{(\sqrt 3 - 1)}^2}} = 2\sqrt 2 $
$\tan \alpha = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{1 - (1/\sqrt 3 )}}{{1 + (1/\sqrt 3 )}} = \tan \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$
$\Rightarrow \alpha = \frac{\pi }{{12}}$
The given equation reduces to
$2\sqrt 2 \cos (\theta - \alpha ) = 2 $
$\Rightarrow \cos \left( {\theta - \frac{\pi }{{12}}} \right) = \cos \frac{\pi }{4}$
$ \Rightarrow $ $\theta - \frac{\pi }{{12}} = 2n\pi \pm \frac{\pi }{4} $
$\Rightarrow \theta = 2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$.
But this value does not satisfy the given equation.
Hence option $(a)$ is the correct answer.
$f(0) = \frac{3}{2} > 0$
$f\left( {\frac{\pi }{2}} \right) = 0 - \frac{\pi }{2} + \frac{1}{2} = \frac{{1 - \pi }}{2} < 0$,
$\left( \because {\;\pi = \frac{{22}}{7}{\rm{ nearly}}} \right)$
$\therefore$ One root lies in the interval $\left[ {0,\,\frac{\pi }{2}} \right]$.
==> ${({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$
==> ${\sin ^2}2x - 2\sin 2x - 2 - 2\alpha = 0$
Let $sin 2x = y$. Then the given equation becomes
${y^2} - 2y - 2(1 + \alpha ) = 0$,
where $ - 1 \le y \le 1$, $({\rm{ }} - 1 \le \sin 2x \le 1)$
For real, discriminant
$ \ge 0$$ \Rightarrow $$3 + 2\alpha \ge 0$
$ \Rightarrow $ $\alpha \ge - \frac{3}{2}$
Also $ - 1 \le y \le 1 \Rightarrow - 1 \le 1 - \sqrt {3 + 2\alpha } \,\, \le 1$
$ \Rightarrow $ $3 + 2\alpha \le 4 \Rightarrow \alpha \le \frac{1}{2}$.
Thus $ - \frac{3}{2} \le \alpha \le \frac{1}{2}$.
$ \Rightarrow $ $5(2{\cos ^2}\theta - 1) + (1 + \cos \theta ) + 1 = 0$
$ \Rightarrow $ $10{\cos ^2}\theta + \cos \theta - 3 = 0$
$ \Rightarrow $ $(5\cos \theta + 3)\,(2\cos \theta - 1) = 0$
$ \Rightarrow $ $\cos \theta = \frac{1}{2},\,\cos \theta = - \frac{3}{5}$
$\Rightarrow \theta = \frac{\pi }{3},\,\pi - {\cos ^{ - 1}}\left( {\frac{3}{5}} \right)$.
$\Rightarrow$ $\frac{1}{2}({2^{\sin x}} + {2^{\cos x}}) \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $
$ \Rightarrow $ ${2^{\sin x}} + {2^{\cos x}} \ge {2.2^{\frac{{\sin x + \cos x}}{2}}}$
$ \Rightarrow $${2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \frac{{\sin x + \cos x}}{2}}}$
and we know that $\sin x + \cos x \ge - \sqrt 2 $
$\therefore $ ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$, for $x = \frac{{5\pi }}{4}$.
==> $\alpha + \beta = (2n + 1)\frac{\pi }{2},n \in I$
$\therefore$ $\sin (\alpha + 2\beta ) = \sin (2\alpha + 2\beta - \alpha )$
$=\sin {\rm{ }}[(2n + 1){\rm{ }}\pi - \alpha ]$
$ = \sin (\,2n\pi + \pi - \alpha \,)$ = $\sin (\,\pi - \alpha \,)\, = \sin \alpha $.
$ \Rightarrow $ $\tan (\cot x) = \tan \left( {\frac{\pi }{2} - \tan x} \right)$
$ \Rightarrow $ $\cot x = n\pi + \frac{\pi }{2} - \tan x $
$\Rightarrow \cot x + \tan x = n\pi + \frac{\pi }{2}$
$ \Rightarrow $ $\frac{2}{{\sin 2x}} = n\pi + \frac{\pi }{2}$
$\Rightarrow \sin 2x = \frac{2}{{n\pi + \frac{\pi }{2}}} = \frac{4}{{(2n + 1)\pi }}$
$2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x - y) = 2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x + y)$
$\therefore $ Either $\sin \frac{1}{2}(x + y) = 0$ or $\sin \frac{1}{2}x = 0$ or $\sin \frac{1}{2}y = 0$
Thus $x + y = - 1,\,\,x - y = - 1$.
When $x + y = 0,$ we have to reject $x + y = 1$ and check with the options or $x + y = - 1$ and solve it with $x - y = 1$ or $x - y = - 1$
which gives $\left( {\frac{1}{2},\,\, - \frac{1}{2}} \right)$ or $\left( { - \frac{1}{2},\,\frac{1}{2}} \right)$ as the possible solution.
Again solving with $x = 0$, we get $(0,{\rm{ }} \pm 1)$ and solving with $y = 0$, we get $( \pm {\rm{ }}1,\,\,0)$ as the other solution.
Thus we have six pairs of solution for $ x$ and $y$.
then $\sin \,(x + \lambda ) + \cos p\,(x + \lambda ) = \sin x + \cos px,\,\,\forall \,\,x \in R$
Putting $x = 0$ and replace $\lambda $ by $ - \lambda $, we have
$\sin \lambda + \cos p\lambda = 1$ and $ - \sin \lambda + \cos p\lambda = 1$
Solving these, we get $\sin \lambda = 0$ so $\lambda = n\pi $ and
$\cos p\lambda = 1$ so $p\lambda = 2m\pi .$
As $\lambda \ne 0,\,\,m$ and $n$ are non-zero integers.
Hence $p = \frac{{2m\pi }}{\lambda },$ which is rational.
$f(x) = \frac{{\sin \,2x}}{{\sin \,\left( {\frac{x}{2}} \right)}} = \frac{{4\,\sin \,\left( {\frac{x}{2}} \right)\,\cos \,\left( {\frac{x}{2}} \right)\,\cos x}}{{\sin \,\left( {\frac{x}{2}} \right)}}$
$ = 4\,\cos \,\left( {\frac{x}{2}} \right)\,\cos x$
The period of $\cos x = 2\pi $ and that of $\cos \frac{x}{2}$ is $4\pi ,$ so period of
$\frac{{\sin \,\,2x}}{{\sin \,\left( {\frac{x}{2}} \right)}}$ is $4\pi .$
For $n = 3,$
$\frac{{\sin \,\left\{ {3\,(x + 4\pi )} \right\}}}{{\sin \,\left\{ {\frac{{(x + 4\pi )}}{3}} \right\}}} = \frac{{\sin \,3x}}{{\sin \,\left( {\frac{x}{3} + \frac{{4\pi }}{3}} \right)}} \ne \frac{{\sin \,3x}}{{\sin \,\left( {\frac{x}{3}} \right)}}$
So, $4\pi $ is not the period for $n = 3.$
Similarly, we can see that $4\pi $ is not the period of $\frac{{\sin \,nx}}{{\sin \left( {\frac{x}{n}} \right)}}$ for $n = 4$ and $5$ also.
==> Period $ = \frac{{2\pi }}{2} = \pi $.
$3 \,sin\, \theta - 4sin^3\theta = 4sin\theta \,sin\, 2\theta \,sin4\theta$
hence either $sin\, \theta = 0 \Rightarrow\, \theta = n\pi$
or $3 - 4sin^2\theta = 4\, sin\, 2\theta \,sin \,4\theta$
$3 - 2 (1 - cos\, 2\theta ) = 2 (cos\, 2\theta - cos \,6\theta )$
or $1 = - 2 cos\, 6\theta$
$cos \,6\theta = - \frac{1}{2} = cos\, \frac{2 \pi}{3}$
$6\theta = 2n \pi ± \frac{2 \pi}{3}$
if $0\, \le\, \theta\, \le\, \pi$ then total solution are $0,\frac{\pi }{9},\frac{{2\pi }}{9},\frac{{4\pi }}{9},\frac{{5\pi }}{9},\frac{{7\pi }}{9},\frac{{8\pi }}{9},\pi $ is $8$ real solutions.
$cos\theta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$ $=$$\frac{{\sqrt 2 \, \pm \,\sqrt 6 }}{4}$
$cos\theta =$ $\frac{{\sqrt 6 + \sqrt 2 }}{4}$ $ \Rightarrow \,\,\theta \,\, = \,\,\frac{\pi }{{12}}\,;\,2\pi \, - \frac{\pi }{{12}}\, = \,\frac{{23\pi }}{{12}}$
$cos\theta =$ $ - \,\frac{{\sqrt 6 - \sqrt 2 }}{4}$
$cos\theta = cos(\pi -5\pi /12) ; cos(\pi +5\pi /12)$
$\theta = 7\pi /12 ; 17\pi /12$
Now apply $C/D$ and then proceed
$\Rightarrow \,sin\theta = 1$
$\Rightarrow \theta = 2n\pi + \pi /2$
$\Rightarrow$ infinite roots
$tan \,(5\pi \,cos\, \theta ) = tan\, \left( {\frac{\pi }{2}\, - \,5\pi \sin \theta } \right)$
$5\pi \,cos\theta = n\pi + \pi /2 - 5\pi \,sin\theta$
$(cos\theta + sin\theta ) = \left( {\frac{{2n + 1}}{{10}}} \right)$
$\Rightarrow -1 < \frac{{2n + 1}}{{10\sqrt 2 }} < 1$
$\Rightarrow$ $ - \,\frac{{10\sqrt 2 \, - 1}}{2}$ $< n <$ $\frac{{10\sqrt 2 \, - 1}}{2}$
$\therefore$ $n= 14$ for each $‘n’$ there are two values of $\theta$
no. of solutions $= 28$
i.e. $2sin^2x - a\, sinx + 2a - 8 = 0$
$sinx = \frac{{a \pm \sqrt {{a^2} - 8(2a - 8)} }}{4} = \frac{{a \pm (a - 8)}}{4}$
$sinx = \frac{{a - 4}}{2}\,or\,2$
Hence $-1 \leq (a- 4)/ 2 \leq 1$
$\Rightarrow$ the range of $a$
$2 \,sin3x \, cos2x = 2\, sin3x \,cosx$
$2sin3x [ cos2x - cosx] = 0$
On solving we get $x = n\pi /3$
$sin^2x -\frac{{\sqrt 3 \, + 1}}{2} \,sinx + \frac{{\sqrt 3 }}{4} = 0$
$4 sin^2x - 2 \sqrt 3 sinx - 2sinx + \sqrt 3 = 0$
On solving we get
$sinx = 1/2 ; \frac{{\sqrt 3 }}{2}$
= $(\pi /6 , 5\pi /6 ; \pi /3 , 2\pi /3 ]$
$\begin{array}{l} - \frac{{3\pi }}{4} \le \frac{x}{2} - \frac{\pi }{2} \le \frac{{3\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{\pi }{2},0,\frac{\pi }{2},\pi ,\frac{{3\pi }}{2},2\pi ,\frac{{5\pi }}{2}\\ - \frac{\pi }{4} \le \frac{x}{2} \le \frac{{5\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \frac{x}{2} - \cos \frac{x}{2} = {(\sin \frac{x}{2} - \cos \frac{x}{2})^2}\\ - \frac{\pi }{2} \le x \le \frac{{5\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,factors\,\sin \frac{x}{2} - \cos \frac{x}{2} = \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\sin \frac{x}{2} - \cos \frac{x}{2} = 1\end{array}$
only circled angle satisfy one of the above equation when $n = 1, 2, 4, 5$
$a_{1}+\frac{a_{3}}{2}=0$
$a_{2}-\frac{a_{3}}{2}=0$
These two equations can have infinite roots.
$2\, cos \,3x \,cos \,2x + 2\, cos \,3x \,cos\, x + cos \,3x = 0$
$cos\, 3x [2 \,cos \,2x + 2 \,cos\, x - 1] = 0$
$\left. \begin{array}{l}x = (2n - 1)\frac{\pi }{6} \Rightarrow \frac{\pi }{6},\frac{{3\pi }}{6},\frac{{5\pi }}{6} = 3\\{2^{nd}} \,\, equation\,\, gives \,\,cos{\mkern 1mu} x = \frac{{1 \pm \sqrt 2 }}{4} = 2\end{array} \right] = 5$
$\Rightarrow tan\, x + tan\, 2x = 0$
$\therefore$ $tan\, 2x = tan\, (- x)$
$2x = n\pi - x$
$x =$ $\frac{{n\pi }}{3}$, $n \in I$
$\theta = n\pi +(-1)^n\alpha$
$n = 0$ $\theta = \alpha$ $sin\theta /3 = sin\alpha /3$
$n = 1$ $\theta = \pi -\alpha$ $sin\theta /3 = sin(\pi /3-\alpha /3)$
$n = -1$ $\theta = -\pi -\alpha$ $sin\theta /3 = sin(-\pi /3-\alpha /3)= -sin(\pi /3+\alpha /3)$
$=\frac{1}{2}+\frac{1}{2 \sin \frac{x}{2}}\left(\sin \frac{9 x}{2}-\sin \frac{x}{2}\right)=0$
$=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}=0 $
$ \Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{9}, \mathrm{n} \neq 9 \mathrm{mm} \in \mathrm{I}$
$=\frac{(2 \sin x-1)(\sin x-1)}{\sin x} $
$ \Rightarrow \mathop {\mathop {\sin x = \frac{1}{2}}\limits_ \Downarrow }\limits_{4\,solutions} {\rm{ }}\,\,\,\,\,{\rm{or }}\,\,\,\,\mathop {\mathop {\,\frac{1}{2}{{\cos }^2}2x}\limits_{ \ge 0} = \mathop {\sin x - 1}\limits_{ \le 0} }\limits_{{\rm{Hence no solution }}} $
$ \Rightarrow \,\left[ {\frac{\pi }{4},\frac{{3\pi }}{4}} \right]$
$(\sin \theta+\cos \theta)^{2}>\frac{3}{2}$
$\sin 2 \theta>\frac{1}{2}$
$2 n \pi+\frac{\pi}{6}<2 \theta<2 n \pi+\frac{5 \pi}{6}$
$n\pi + \frac{\pi }{{12}} < \theta < n\pi + \frac{{5\pi }}{{12}}$
$5\left(\frac{1+\cos 2 \theta}{2}\right)-3\left(\frac{1-\cos 2 \theta}{2}\right)+3 \sin 2 \theta=7$
$4 \cos 2 \theta+3 \sin 2 \theta=6$
but $4 \cos 2 \theta+3 \sin 2 \theta \leq \sqrt{4^{2}+3^{2}}=5$
$\therefore $ Solution does not exist.