Question

If $2{\sin }^{2}x=3\cos x$, where $0\le x\le 2\mathrm{\pi }$, then find the value of x.

Answer 1

The given equation is $2{\sin }^{2}x=3\cos x$.
Now,
$$\begin{gathered} 2{\sin }^{2}x=3\cos x \\ \Rightarrow 2\left(1-{\cos }^{2}x\right)=3\cos x \\ \Rightarrow 2{\cos }^{2}x+3\cos x-2=0 \\ \Rightarrow \left(2\cos x-1\right)\left(\cos x+2\right)=0 \end{gathered}$$
$\Rightarrow \cos x=\frac{1}{2}\mathrm{or}\cos x=-2$
But, $\cos x=-2$ is not possible. $\left(-1\le \cos x\le 1\right)$
$$\begin{gathered} \therefore \cos x=\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathbf{Z}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(\cos x=\cos \alpha \Rightarrow x=2n\mathrm{\pi }\pm \alpha ,n\in \mathbf{Z}\right) \end{gathered}$$
Putting n = 0 and n = 1, we get
$x=\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\left(0\le x\le 2\mathrm{\pi }\right)$