If $2 \sin ^{2} x=3 \cos x$, where $0 \leq x \leq 2 \pi$, then find the value of $x$.
The given equation is $2 \sin ^{2} x=3 \cos x$.
Now,
$2 \sin ^{2} x=3 \cos x$
$\Rightarrow 2\left(1-\cos ^{2} x\right)=3 \cos x$
$\Rightarrow 2 \cos ^{2} x+3 \cos x-2=0$
$\Rightarrow(2 \cos x-1)(\cos x+2)=0$
$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-2$
But, $\cos x=-2$ is not possible. $\quad(-1 \leq \cos x \leq 1)$
$\therefore \cos x=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow x=2 n \pi \pm \frac{\pi}{3}, n \in \mathbf{Z} \quad(\cos x=\cos \alpha \Rightarrow x=2 n \pi \pm \alpha, n \in \mathbf{Z})$
Putting $n=0$ and $n=1$, we get
$x=\frac{\pi}{3}, \frac{5 \pi}{3} \quad(0 \leq x \leq 2 \pi)$
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