Answer
For the provided equation, the value of x is $ x=0,\frac{\pi }{3},\frac{5\pi }{3}$.
Work Step by Step
We have $2{{\sin }^{2}}x+3\cos x-3=0$
By solving,
$\begin{align}
& 2{{\sin }^{2}}x+3\cos x-3=0 \\
& 2\left( 1-{{\cos }^{2}}x \right)+3\cos x-3=0 \\
& 2-2{{\cos }^{2}}x+3\cos x-3=0 \\
& 2{{\cos }^{2}}x-3\cos x+1=0
\end{align}$
And factorizing:
$\begin{align}
& 2{{\cos }^{2}}x-2\cos x-\cos x+1=0 \\
& 2\cos x\left( \cos x-1 \right)-1\left( \cos x-1 \right)=0 \\
& \left( 2\cos x-1 \right)\left( \cos x-1 \right)=0 \\
& \cos x=1,\frac{1}{2}
\end{align}$
And for
$\begin{align}
& \cos x=1 \\
& x=0
\end{align}$
And for
$\begin{align}
& \cos x=\frac{1}{2} \\
& x=\frac{\pi }{3},\frac{5\pi }{3}
\end{align}$
Hence, $ x=0,\frac{\pi }{3},\frac{5\pi }{3}$.
