Question

sin(pi/2-x)+2cos2x=1 решить уравнение

Answer 1

sin(pi/2 - x) = cosx
cosx + 2cos2x = 1
2cos²(x) - 2sin²(x) + cosx = 1
sin^2(x) = 1-cos^2(x)
2cos²(x) - 2 + 2cos²(x) + cosx = 1
t = cosx
4t² + t - 3 = 0
4(t + 3/2)(t - 1/2) = 0
[t = -3/2] Посторонний корень, тк cosx ∈ [-1;1]
t = 1/2
cosx = 1/2
x = ±π/3 + 2πk, k ∈ Ζ