Statement-1: The number of common solutions of the trigonometric equations $2 \sin ^2 \theta$ $-\cos 2…
Statement-1: The number of common solutions of the trigonometric equations $2 \sin ^2 \theta$ $-\cos 2 \theta=0$ and $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is two.
Statement-2: The number of solutions of the equation, $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0, \pi]$ is two.
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
Statement-1 is false; Statement-2 is true.
Statement-1 is true; Statement-2 is false.
Solution
$2 \sin ^2 \theta-\cos 2 \theta=0$
$\Rightarrow 2 \sin ^2 \theta-\left(1-2 \sin ^2 \theta\right)=0$
$\Rightarrow 2 \sin ^2 \theta-1+2 \sin ^2 \theta=0$
$\Rightarrow 4 \sin ^2 \theta=1 \Rightarrow \sin \theta=\pm \frac{1}{2}$
$\therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}, \theta \in[0,2 \pi]$
$\therefore \theta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$
$
\begin{aligned}
& \text { Now } 2 \cos ^2 \theta-3 \sin \theta=0 \\
\Rightarrow & 2\left(1-\sin ^2 \theta\right)-3 \sin \theta=0 \\
\Rightarrow & -2 \sin ^2 \theta-3 \sin \theta+2=0 \\
\Rightarrow & -2 \sin ^2 \theta-4 \sin \theta+\sin \theta+2=0 \\
\Rightarrow & 2 \sin ^2 \theta-\sin \theta+4 \sin \theta-2=0 \\
\Rightarrow & \sin \theta(2 \sin \theta-1)+2(2 \sin \theta-1)=0 \\
\Rightarrow & \sin \theta=\frac{1}{2},-2
\end{aligned}
$
But $\sin \theta=-2$, is not possible
$
\therefore \quad \sin \theta=\frac{1}{2}, \Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6}
$
Hence, there are two common solution, there each of the statement-1 and 2 are true but statement- 2 is not a correct explanation for statement-1