Statement-I: In the interval [0, 2pi],the num | vedclass

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(A) For Statement-$I$: Equation $1$: $2 \sin^2 	heta - \cos 2	heta = 0$ Using $\cos 2	heta = 1 - 2 \sin^2 	heta$,we get $2 \sin^2 	heta - (1 - 2

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Statement-$I$ and Statement-$II$ are both true

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(A) For Statement-$I$: Equation $1$: $2 \sin^2 	heta - \cos 2	heta = 0$ Using $\cos 2	heta = 1 - 2 \sin^2 	heta$,we get $2 \sin^2 	heta - (1 - 2 \sin^2 	heta) = 0 \implies 4 \sin^2 	heta = 1 \implies \sin^2 	heta = \frac{1}{4} \implies \sin 	heta = \pm \frac{1}{2}$. In $[0, 2\pi]$,$	heta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$. Equation $2$: $2 \cos^2 	heta - 3 \sin 	heta = 0$ Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $2(1 - \sin^2 	heta) - 3 \sin 	heta = 0 \implies 2 - 2 \sin^2 	heta - 3 \sin 	heta = 0 \implies 2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$. $(2 \sin 	heta - 1)(\sin 	heta + 2) = 0$. Since $\sin 	heta 
eq -2$,we have $\sin 	heta = \frac{1}{2}$. In $[0, 2\pi]$,$	heta = \frac{\pi}{6}, \frac{5\pi}{6}$. Common solutions are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$,so there are two common solutions. Statement-$I$ is true. For Statement-$II$: The solutions of $2 \cos^2 	heta - 3 \sin 	heta = 0$ in $[0, \pi]$ are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5\pi}{6}$. There are two solutions. Statement-$II$ is true.

Statement-$I$: In the interval $[0, 2\pi]$,the number of common solutions of the equations $2 \sin^2 \theta - \cos 2\theta = 0$ and $2 \cos^2 \theta - 3 \sin \theta = 0$ is two.
Statement-$II$: The number of solutions of $2 \cos^2 \theta - 3 \sin \theta = 0$ in $[0, \pi]$ is two.

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Statement-$I$: In the interval $[0, 2\pi]$,the number of common solutions of the equations $2 \sin^2 \theta - \cos 2\theta = 0$ and $2 \cos^2 \theta - 3 \sin \theta = 0$ is two. Statement-$II$: The number of solutions of $2 \cos^2 \theta - 3 \sin \theta = 0$ in $[0, \pi]$ is two.