Решение !!!!!!!!!!!!!!

Решите задачу:

$4sin^2x=\sqrt2tgx\; ,\; \; ODZ:\; x\ne \frac{\pi}{2}+\pi n,\; n\in Z\\\\4sin^2x-\frac{\sqrt2sinx}{cosx}=0\\\\ \frac{4sin^2x\cdot cosx-\sqrt2sinx}{cos}=0\\\\ \frac{\sqrt2sinx\cdot (2\sqrt2sinx\cdot cosx-1)}{cosx} =0\; \; \to \; \; \left \{ {{\sqrt2sinx\cdot (2\sqrt2sinx\cdot cosx)=0} \atop {cosx\ne 0}} \right. \\\\a)\; \; sinx=0\; ,\; \; x=\pi k,\; k\in Z\\\\b)\; \; 2\sqrt2sinx\cdot cosx-1=0\\\\\sqrt2\cdot sin2x-1=0\\\\sin2x=\frac{1}{\sqrt2}\; ,\; \; \; \; \frac{1}{\sqrt2}=\frac{\sqrt2}{2}$

$2x=(-1)^{m}\cdot \frac{\pi }{4}+\pi m,\; m\in Z\\\\x=(-1)^{m}\cdot \frac{\pi}{8}+\frac{\pi m}{2},\; m\in Z\\\\Otvet:\; \; x=\pi k,\; k\in Z\; \; ;\; \; x=(-1)^{m}\frac{\pi}{8}+\frac{\pi m}{2},\; m\in Z \; .$