Question

$sin^{2}x$ - $cos^{2}x$ = cos4x

Answer 1

sin²x-cos²x=cos4x
-(cos²x-sin²x)=cos4x
-cos2x-cos4x=0
cos2x+cos4x=0
2cos(²ˣ⁺⁴ˣ/₂)cos(²ˣ⁻⁴ˣ/₂)=0
cos3x cos(-x)=0
cos3x cosx=0

a) cos3x=0
    3x=π/2 + πn
    x= π/6 + (πn)/3, n∈Z

б) сosx=0
    x=π/2 + πn, n∈Z

Ответ: π/6 + (πn)/3, n∈Z;
            π/2 + πn, n∈Z.