Question

The number of solution(s) of $2{\sin }^2\theta -\cos 2\theta =0$ and $2{\cos }^2\theta -3\sin \theta =0$ for $\theta \in [0,2\pi ]$ is

Answer 1

Given : $2{\sin }^2\theta -\cos 2\theta =0$ and $2{\cos }^2\theta -3\sin \theta =0$

Now,
$2{\sin }^2\theta -\cos 2\theta =0$
$\Rightarrow 2{\sin }^2\theta -(1-2{\sin }^2\theta )=0\Rightarrow 4{\sin }^2\theta =1\Rightarrow {\sin }^2\theta =\frac{1}{4}\Rightarrow \sin \theta =\pm \frac{1}{2}\cdots \left(1\right)$

Also,
$2{\cos }^2\theta -3\sin \theta =0\Rightarrow 2(1-{\sin }^2\theta )-3\sin \theta =0\Rightarrow 2{\sin }^2\theta +3\sin \theta -2=0\Rightarrow (2\sin \theta -1)(\sin \theta +2)=0\Rightarrow \sin \theta =\frac{1}{2}(\because \sin \theta \in [-1,1\left]\right)\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$\sin \theta =\frac{1}{2}\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6}(\because \theta \in [0,2\pi \left]\right)$