Вопрос в картинках...

Решите задачу:

$\sqrt{sinx\cdot cosx}=-cosx\; ,\; \; ODZ:\; \left \{ {{sinx\cdot cosx \geq 0} \atop {cosx \leq 0}} \right. \; \to \; \left \{ {{sinx \leq 0} \atop {cosx \leq 0}} \right. \; \to \\\\x\in 3 \; chetvert\; ,\; x\in [\frac{\pi}{2}+2\pi n,\pi +2\pi n]\\\\sinx\cdot cosx=cos^2x\\\\cosx(cosx-sinx)=0\\\\a)\; \; cosx=0,\; x=\frac{\pi}{2}+\pi n,\; +ODZ\; \to \; x=\frac{\pi}{2}+2\pi n,\; n\in Z\\\\b)\; \; cosx-sinx=0\, |:cosx\ne 0\\\\1-tgx=0\\\\tgx=1\\\\x=\frac{\pi}{4}+\pi n,\; n\in Z$

$\left \{ {{x=\frac{\pi}{4}+\pi n} \atop {x\in [\frac{\pi}{2}+2\pi n;\pi +2\pi n]}} \right. \; \to \; net\; reshenij$