Question

Решите уравнение, пожалуйста

Answer 1

$sin(x+\frac{\pi}{6})sin(x-\frac{\pi}{6})=\frac{1}{4}<=>\frac{1}{4}-\frac{cos(2x)}{2}=\frac{1}{4}<=>\\<=>\frac{cos(2x)}{2}=0<=>cos(2x)=0=>x=\frac{\pi}{4}+\frac{\pi k}{2}$

k∈Z