The number of solutions to sin left(pi sin ^2 | vedclass

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Question

(d)

The given equation

$\sin \left(\pi \sin ^2 	hetaight)+\sin \left(\pi \cos ^2 	hetaight)=2 \cos \left(\frac{\pi}{2} \cos 	hetaight)$

Vedclass · Accepted Answer

$7$

Vedclass · Answer

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

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 $0 \leq \theta \leq 2 \pi$ is

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