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

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(D) The given equation is $\sin(\pi \sin^2 	heta) + \sin(\pi \cos^2 	heta) = 2 \cos(\frac{\pi}{2} \cos 	heta)$.Using the identity $\sin A + \sin B

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$7$

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(D) The given equation is $\sin(\pi \sin^2 	heta) + \sin(\pi \cos^2 	heta) = 2 \cos(\frac{\pi}{2} \cos 	heta)$.Using the identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:$2 \sin(\frac{\pi(\sin^2 	heta + \cos^2 	heta)}{2}) \cos(\frac{\pi(\sin^2 	heta - \cos^2 	heta)}{2}) = 2 \cos(\frac{\pi}{2} \cos 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$ and $\sin^2 	heta - \cos^2 	heta = -\cos 2	heta$,we have:$2 \sin(\frac{\pi}{2}) \cos(-\frac{\pi}{2} \cos 2	heta) = 2 \cos(\frac{\pi}{2} \cos 	heta)$.Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(-x) = \cos x$,this simplifies to:$\cos(\frac{\pi}{2} \cos 2	heta) = \cos(\frac{\pi}{2} \cos 	heta)$.This implies $\frac{\pi}{2} \cos 2	heta = 2n\pi \pm \frac{\pi}{2} \cos 	heta$,so $\cos 2	heta = 4n \pm \cos 	heta$ for $n \in Z$.Case $I$: $\cos 2	heta - \cos 	heta = 4n$. For $n=0$,$2\cos^2 	heta - 1 - \cos 	heta = 0$,so $(2\cos 	heta + 1)(\cos 	heta - 1) = 0$. Thus $\cos 	heta = 1$ or $\cos 	heta = -1/2$. Solutions in $[0, 2\pi]$ are $	heta = 0, 2\pi, 2\pi/3, 4\pi/3$.Case $II$: $\cos 2	heta + \cos 	heta = 4n$. For $n=0$,$2\cos^2 	heta + \cos 	heta - 1 = 0$,so $(2\cos 	heta - 1)(\cos 	heta + 1) = 0$. Thus $\cos 	heta = 1/2$ or $\cos 	heta = -1$. Solutions in $[0, 2\pi]$ are $	heta = \pi, \pi/3, 5\pi/3$.Total solutions are ${0, 2\pi, 2\pi/3, 4\pi/3, \pi, \pi/3, 5\pi/3}$,which is $7$ solutions.

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

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