[tex]\bf cos(2x)=-\cfrac{\sqrt{2}}{2}\implies cos^{-1}[cos(2x)]=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)
\\\\\\
\measuredangle 2x=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)\implies \measuredangle 2x=
\begin{cases}
\frac{3\pi }{4}\\\\
\frac{5\pi }{4}
\end{cases}\\\\
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x=\cfrac{3\pi }{2\cdot 4}\implies \measuredangle x=\cfrac{3\pi }{8}\\\\
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x=\cfrac{5\pi }{2\cdot 4}\implies \measuredangle x=\cfrac{5\pi }{8}[/tex]
