sin(pi/2-x)+2cos2x=1 решить уравнение

sin(pi/2 - x) = cosxcosx + 2cos2x = 12cos²(x) - 2sin²(x) + cosx = 1sin^2(x) = 1-cos^2(x)2cos²(x) - 2 + 2cos²(x) + cosx = 1t = cosx4t² + t - 3 = 04(t + 3/2)(t - 1/2) = 0[t = -3/2] Посторонний корень, тк cosx ∈ [-1;1]t = 1/2cosx = 1/2x = ±π/3 + 2πk, k ∈ Ζ