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

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

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A)
$2\sin^2\theta-\cos 2\theta=0$
$2\sin^2\theta-(1-2\sin^2\theta)=0$
$4\sin^2\theta=1$
$\sin\theta=\large\frac{1}{2},$$\sin\theta=-\large\frac{1}{2}$
$2\cos^2\theta-3\sin\theta=0$
$2-2\sin^2\theta-3\sin\theta=0$
$2\sin^2\theta+3\sin\theta-2=0$
$(2\sin\theta-1)(\sin\theta+2)=0$
$\sin\theta=\large\frac{1}{2}$
$\sin\theta\neq -2$
Hence we get $\sin\theta=\large\frac{1}{2}$
$\Rightarrow \theta=\large\frac{\pi}{6},\frac{5\pi}{6}$
Hence (c) is the correct answer.