Помогите решить cos^2 3x=1/2

[latex]cos^23x=\frac{1}2\\cos3x=б\frac{\sqrt2}2\\\\ \left[\begin{array}{ccc}cos3x=\frac{\sqrt2}2\\cos3x=-\frac{\sqrt2}2\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}3x=б\frac{\pi}4+2\pi n;n\in Z\\3x=б\frac{3\pi}4+2\pi n;n\in Z\end{array}\right=\ \textgreater \ \\\\\\=\ \textgreater \ 3x=б\frac{\pi}4+\pi n;n\in Z=\ \textgreater \ x=б\frac{\pi}{12}+\frac{\pi n}3;n\in Z[/latex]