Question

Решите тригонометрическое уравнение

Answer 1

$\sqrt{sinxcosx}=-cosx\\ (\sqrt{sinxcosx})^{2}=(-cosx)^{2}\\sinxcosx=cos^{2}x\\ \frac{sinxcosx}{cos^{2}x}= \frac{cos^{2}x}{cos^{2}x} \\ \frac{sinx}{cosx}=1\\tgx=1\\x= \frac{\pi}{4}+\pi n,n\in Z$