The number of solutions of the pair of equati | vedclass

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(C) Given equations are $2 \sin^2 	heta - \cos 2 	heta = 0$ and $2 \cos^2 	heta - 3 \sin 	heta = 0$. From the first equation,$2 \sin^2 	heta - (1

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two

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(C) Given equations are $2 \sin^2 	heta - \cos 2 	heta = 0$ and $2 \cos^2 	heta - 3 \sin 	heta = 0$. From the first equation,$2 \sin^2 	heta - (1 - 2 \sin^2 	heta) = 0$,which simplifies to $4 \sin^2 	heta = 1$,so $\sin^2 	heta = \frac{1}{4}$,implying $\sin 	heta = \pm \frac{1}{2}$. From the second equation,$2(1 - \sin^2 	heta) - 3 \sin 	heta = 0$,which is $2 - 2 \sin^2 	heta - 3 \sin 	heta = 0$,or $2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$. Factoring the quadratic,$(2 \sin 	heta - 1)(\sin 	heta + 2) = 0$. Since $\sin 	heta$ cannot be $-2$,we must have $\sin 	heta = \frac{1}{2}$. Comparing both conditions,the common solutions satisfy $\sin 	heta = \frac{1}{2}$. In the interval $[0, 2 \pi]$,$\sin 	heta = \frac{1}{2}$ at $	heta = \frac{\pi}{6}$ and $	heta = \frac{5 \pi}{6}$. Thus,there are $2$ solutions.

The number of solutions of the pair of equations $2 \sin^2 \theta - \cos 2 \theta = 0$ and $2 \cos^2 \theta - 3 \sin \theta = 0$ in the interval $[0, 2 \pi]$ is

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