Question

Cos4xcos2x+sin4xsin2x>0,5

Answer 1

$\mathtt{cos4xcos2x+sin4xsin2x\ \textgreater \ 0,5}\\\mathtt{cos(4x-2x)\ \textgreater \ 0,5}\\\mathtt{cos2x\ \textgreater \ 0,5}\\\mathtt{ -\dfrac{\pi}{3}\ \textless \ 2x\ \textless \ \dfrac{\pi}{3} }\\\mathtt{ -\dfrac{\pi}{3}+2\pi k\ \textless \ 2x\ \textless \ \dfrac{\pi}{3}+2\pi k,k\in\mathbb{Z}}\\\mathrb{x\in(- \dfrac{\pi}{6}+\pi k; \dfrac{\pi}{6}+\pi k),k\in\mathbb{Z}}$

Answer 2

Cos4xcos2x+sin4xsin2x>0,5
cos(4x-2x)>0,5
cos2x>0,5
-π/3+2πk<2x<π/3+2πk
-π/6+πk<x<π/6+πk,k∈z      
Ответ x∈(-π/6+πk;π/6+πk,k∈z)