Question

(4cos^2x-3)/(log_5(-tgx)=0

CPO4HO

Answer 1

$frac{4cos^2x-3}{log_5(-tgx)}=0$

$left { {{log_5(-tgx)neq0} atop {4cos^2x-3=0}} right.$ 

$begin{cases} -tgx>0\-tgxneq1\cos^2x=frac{3}{4} end{cases}$ 

$begin{cases} xneq-frac{pi}{4}\x = frac{pi}{6}+2pi kend{cases}$

$x = frac{pi}{6}+2pi k,$ k целое