Question

The number of solutions to $\sin \left(\pi {\sin }^2\theta \right)+\sin \left(\pi {\cos }^2\theta \right)=2\cos \left(\frac{\pi }{2}\cos \theta \right)$ satisfying $0\le 0\le 2\pi$ is

• A. 1
• B. 2
• C. 4
• D. 7

Answer 1

Answer: (D)

The given equation
$\sin (\pi {\sin }^2\theta )+\sin (\pi {\cos }^2\theta )$
$=2\cos \left(\frac{\pi }{2}\cos \theta \right)$
$\Rightarrow 2\sin \left(\frac{\pi \left({\sin }^2\theta +{\cos }^2\theta \right)}{2}\right)$
$\cos \left(\frac{\pi \left({\sin }^2\theta -{\cos }^2\theta \right)}{2}\right)=2\cos \left(\frac{\pi }{2}\cos \theta \right)$
$\Rightarrow \cos \left(\frac{\pi }{2}\cos 2\theta \right)$
$=\cos \left(\frac{\pi }{2}\cos \theta \right)$
$\Rightarrow \frac{\pi }{2}\cos 2\theta =2n\pi \pm \frac{\pi }{2}\cos \theta ,n\in$ Integers
$\Rightarrow \cos 2\theta =4n\pm \cos \theta ,n\in I$
Case I
If $\cos 2\theta =4n+\cos \theta$
$\Rightarrow \cos 2\theta -\cos \theta =4n$
The above equation will hold only if $n=0$,
so $\cos 2\theta =\cos \theta$
$\Rightarrow 2{\cos }^2\theta -\cos \theta -1=0$
$\Rightarrow \cos \theta =1,-\frac{1}{2}$
$\Rightarrow \theta =2k\pi ,2k\pi \pm \frac{2\pi }{3},k\in I$
$\because \theta \in \left[0,2\pi \right]$
So, $\theta =0,2\pi ,\frac{2\pi }{3},\frac{4\pi }{3}$
Case II
If $\cos 2\theta =4n-\cos \theta$
$\Rightarrow \cos 2\theta +\cos \theta =4n$
The above equation will hold only if $n=0$,
so $\cos 2\theta +\cos \theta =0$
$\Rightarrow 2{\cos }^2\theta +\cos \theta -1=0$
$\Rightarrow \cos \theta =-1,\frac{1}{2}$
$\Rightarrow \theta =\left(2m+1\right)\pi ,2m\pi \pm \frac{\pi }{3},m\in I$
$\because \theta \in \left[0,2\pi \right]$
So, $\theta =\pi ,\frac{\pi }{3},\frac{5\pi }{3}$
$\therefore$ There are total $7$ solutions