Question

logsinx(4cos2x-1)=0 срочно надо памагити

Answer 1

$\log_{\sin x}(4\cos2x-1)=0\\\log_a(1)=0\\\4cos2x-1=1\\\cos2x=0.5\\x=\pm\frac{\pi}{6} +\pi k , k\in\mathbb{Z}$