Question

(4cos^2x-3)/(log_5(-tgx)=0CPO4HO

Answer 1

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

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

$\begin{cases} -tgx>0\\-tgx\neq1\\cos^2x=\frac{3}{4} \end{cases}$ 

$\begin{cases} x\neq-\frac{\pi}{4}\\x = \frac{\pi}{6}+2\pi k\end{cases}$

$x = \frac{\pi}{6}+2\pi k,$ k целое