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
$\sin (\pi \, \sin^{2}\, \theta) + \sin (\pi \, \cos^{2}\,\theta)$
$=2\, \cos \left(\frac{\pi}{2}\cos\theta\right)$
$\Rightarrow 2 \sin \left(\frac{\pi\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{2}\right)$
$\cos \left(\frac{\pi\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}{2}\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$
$\Rightarrow \cos \left(\frac{\pi}{2}\cos\,2\theta\right)$
$=\cos \left(\frac{\pi}{2} \cos\theta\right)$
$\Rightarrow \frac{\pi}{2}\cos\,2\theta=2\,n\pi \pm \frac{\pi}{2} \cos\theta, n \in$ 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