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

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;4$

Given equations are
$2\sin^2\theta-\cos 2\theta=0$-----(1)
$2\cos^2\theta-3\sin\theta=0$-----(2)
Adding (1) & (2) we get,
$2=\cos 2\theta+3\sin\theta$
$2\sin^2\theta-3\sin\theta+1=0$
$\sin\theta=1$
$\cos 2\theta=2$
Hence rejected.
$\sin\theta=\large\frac{1}{2}$
$\theta=\large\frac{\pi}{6}$ or $\large\frac{5\pi}{6}$
Hence 2 solution is possible.
Hence (c) is the correct answer.