Question

Решить уравнение:
cosx(tgx-cosx)=-sin^2x

Answer 1

$cosx(tgx-cosx)=-sin^2x$
$sinx-cos^2x=-sin^2x$
$sinx-cos^2x+sin^2x=0$
$sinx-1+2sin^2x=0$
$(sinx+1)(2sinx-1)=0$
$sinx+1=0\hspace*{50}2sinx-1=0$
$sinx=1\hspace*{65}2sinx=1$
$x=\frac{\pi}{2}+2{\pi}n;\hspace*{2}n\in{Z}\hspace*{15}sinx=\frac{1}{2}$
$x=\frac{\pi}{2}+2{\pi}n;\hspace*{2}n\in{Z}\hspace*{15}x=(-1)^{n}*arcsin(\frac{1}{2})+\pi{n};\hspace*{2}z\in{Z}$
$x=\frac{\pi}{2}+2{\pi}n;\hspace*{2}n\in{Z}\hspace*{15}x=(-1)^{n}*\frac{\pi}{6}+\pi{n};\hspace*{2}z\in{Z}$