Question

решить задание на фото

Answer 1

$ctg^2x=3\ \ ,\ \ \ \dfrac{cos^2x}{sin^2x}=3\ \ ,\ \ \ \dfrac{1+cos2x}{1-cos2x}=3\ \ ,\\\\\\1+cos2x=3-3cos2x\ \ ,\ \ \ 4cos2x=2\ \ ,\ \ \ cos2x=\dfrac{1}{2}\ ,\\\\\\2x=\pm \dfrac{\pi}{3}+2\pi n\ ,\ n\in Z\\\\\\x=\pm \dfrac{\pi}{6}+\pi n\ ,\ n\in Z$