Question

Решите уравнение:
8cos²x=sin²x

Answer 1

$8cos^2x=sin^2x \ 8cos^2x=1-cos^2x \ 9cos^2x=1 \ cos^2x= dfrac{1}{9} \ left[begin{array}{I} cosx= dfrac{1}{3} \ cosx=- dfrac{1}{3} end{array}}$
$left[begin{array}{I} x= pm arccos( dfrac{1}{3})+2 pi k \ x= pm arccos( -dfrac{1}{3})+2 pi k end{array}}; k in Z$

Ответ: $left[begin{array}{I} x= pm arccos( dfrac{1}{3})+2 pi k \ x= pm arccos( -dfrac{1}{3})+2 pi k end{array}}; k in Z$