Question

уравнение:
cosx = |sinx|

Answer 1

$begin{cases} cos x>0\ cos^2x=sin^2x end{cases}\ begin{cases} cos x>0\ cos^2x-sin^2x=0 end{cases}\ begin{cases} cos x>0\ cos 2x=0 end{cases}\ begin{cases} cos x>0\ 2x=fracpi2+pi n, ninmathbb Z end{cases}\ begin{cases} cos x>0\ x=fracpi4+fracpi2 n end{cases}\ boxed{x=pmfracpi4+2pi n,;ninmathbb Z}$