Answer
The solutions of the provided equation are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},$ and $\frac{7\pi }{4}$
Work Step by Step
Consider the given equation,
$\begin{align}
& 2{{\sin }^{2}}x-1=0 \\
& 2{{\sin }^{2}}x=1 \\
& {{\sin }^{2}}x=\frac{1}{2} \\
\end{align}$
$\sin x=\pm \frac{1}{\sqrt{2}}$
For, $\sin x=\frac{1}{\sqrt{2}}$
$x=\frac{\pi }{4}$
or
$\begin{align}
& x=\pi -\frac{\pi }{4} \\
& =\frac{3\pi }{4}
\end{align}$
And for, $\sin x=-\frac{1}{\sqrt{2}}$
$\begin{align}
& x=\pi +\frac{\pi }{4} \\
& =\frac{5\pi }{4}
\end{align}$
or
$\begin{align}
& x=2\pi -\frac{\pi }{4} \\
& =\frac{7\pi }{4}
\end{align}$
Therefore, the solutions of the provided equation are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},$ and $\frac{7\pi }{4}$.
