2cos^2(2х+пи/3)-sin^2(х+пи/6)=2

Ответ:

Пошаговое объяснение:

2cos²(2x + π/3) - sin²(x + π/6) = 2

2cos²(2x + π/3) - (1 - cos(2x + π/3))/2 = 2

4cos²(2x + π/3) - 1 + cos(2x + π/3) = 4

4cos²(2x + π/3) + cos(2x + π/3) - 5 = 0

Замена: t = cos(2x + π/3), |t| ≤ 1.

4t² + t - 5 = 0

D = 81

t₁ = (-1 - 9)/(2·4) = -5/4 - не удовлетворяет условию |t| ≤ 1

t₂ = (-1 + 9)/(2·4) = 1

cos(2x + π/3) = 1

2x + π/3 = 2πn, n ∈ Z

x = -π/6 + πn, n ∈ Z