Browse Questions

# 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

$(a)zero\qquad(b)\;one\qquad(c)\;two\qquad(d)\;four$

$2\sin^2\theta-\cos 2\theta=0$
$1-2\cos 2\theta=0$
$\cos 2\theta=\large\frac{1}{2}$
$2\theta=\large\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3},\frac{11\pi}{3}$
$\theta=\large\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$------(1)
Where $\theta \in [0,2\pi]$
Also,$2\cos^2\theta-3\sin\theta=0$
$\Rightarrow 2\sin^2\theta+3\sin\theta-2=0$
$\Rightarrow (2\sin\theta-1)(\sin\theta+2)=0$
$\sin\theta=\large\frac{1}{2}$
$\sin\theta\neq -2$
$\theta=\large\frac{\pi}{6},\frac{5\pi}{6}$----(2)
Where $\theta \in [0,2\pi]$
Combining (1) and (2) we get,
$\theta=\large\frac{\pi}{6},\frac{5\pi}{6}$
$\therefore$ Two solutions are there.
Hence (c) is the correct option.