Answer
The solutions of the given equation are $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2}$.
Work Step by Step
The equation $2{{\cos }^{2}}x+3\sin x-3=0$ can be written as,
$\begin{align}
& \left( 2{{\cos }^{2}}x-1 \right)+3\sin x-2=0 \\
& \left( 1-2{{\sin }^{2}}x \right)+3\sin x-2=0 \\
& -2{{\sin }^{2}}x+3\sin x-1=0 \\
& 2{{\sin }^{2}}x-3\sin x+1=0
\end{align}$
By further solving the quadratic equation, we get,
$\begin{align}
& 2{{\sin }^{2}}x-2\sin x-\sin x+1=0 \\
& \left( 2\sin x-1 \right)\left( \sin x-1 \right)=0 \\
& \sin x=\frac{1}{2},1
\end{align}$
Thus, in the given range, we have $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2}$ as the required solutions.
