Cos^2x-sin^2x-sinx=0

Решение Cos²x - sin²x - sinx = 0 1 - sin²x - sin²x - sinx = 0 2sin²x + sinx - 1 = 0 sinx = t 2t² + t - 1 = 0 D = 1 + 4*2*1 = 9 t₁ = (- 1 - 3)/4 t₁ = - 1 t₂ = (- 1 + 3)/4 t₂ = 1/2 1)  sinx = - 1 x₁ = - π/2 + 2πk, k∈Z 2)  sinx = 1/2 x = (-1)^n *arcsin(1/2) + πn, n∈ Z x₂ =  (-1)^n *(π/6) + πn, n∈ Z