Cos^2(x)+cos^2(2x)=1

cos²x+cos²2x=1,  cos²2x=1-cos²x (cos2x)²=sin²x (1-2sin²x)²=sin²x 1-4sin²x+4sin⁴x-sin²x=0 4sin⁴x-5sin²x+1=0 sin²x=t, t∈[-1;1] 4t²-5t+1=0 D=25-16=9 t₁=1, t₂=1/4 t=1 sin²x=1, sinx=+-1 sinx=-1, x=-π/2+2πn, n∈Z sinx=1, x=π/2+2πn, n∈Z t=1/4 sin²x=1/4, sinx=+-1/2 sinx=-1/2, x=(-1)^(n+1)*π/6+πn, n∈Z sinx=1/2, x=(-1)^n*π/6+πn, n∈Z  ответ: x₁=-π/2+2πn, n∈Z              x₂=π/2+2πn, n∈Z              x₃=(-1)^(n+1)*π/6+πn, n∈Z              x₄=(-1)^n*π/6+πn, n∈Z