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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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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


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$2\sin^2\theta-\cos 2\theta=0$
$1-2\cos 2\theta=0$
$\cos 2\theta=\large\frac{1}{2}$
Where $\theta \in [0,2\pi]$
$\Rightarrow 2\sin^2\theta+3\sin\theta-2=0$
$\Rightarrow (2\sin\theta-1)(\sin\theta+2)=0$
$\sin\theta\neq -2$
Where $\theta \in [0,2\pi]$
Combining (1) and (2) we get,
$\therefore$ Two solutions are there.
Hence (c) is the correct option.
answered Oct 17, 2013 by sreemathi.v

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