Помогите решитьcos^2 16x + cos^2 8x = 1

cos²16x+cos²8x=1 cos²16x=(cos(2*8x))²=(2cos²8x-1)²=4cos⁴8x-4cos²8x+1 4cos⁴8c-4cos²8x+cos²8x+1-1=0 4cos⁴8x-3cos²8x=0 cos²8x*(4cos²8x-3)=0  cos²8x=0 или 4cos²8x-3=0 1. cos²8x=0, cos8x=0. 8x=π/2+πn, n∈Z |:8.   x=π/16+πn/8, n∈Z 2. 4cos²8x-3=0, cos²8x=3/4,  cos8x=+-√(3/4) cos8x=-√3/4. 8x=+-arccos(-√3/2)+2πn, n∈Z.   8x=+-(π-arccos(√3/2))+2πn, n∈Z,  8x=+-5π/6+2πn, n∈Z |:8 x=+-5π/48+πn/4, n∈Z cos8x=√3/2, 8x=+-arccos(√3/2)+2πn, n∈Z. 8x=+-π/6+2πn, n∈Z |:8 x=+-π/48+πn/4, n∈Z