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.
(x+2)(x+3)=6
x^2+3x+2x+6=6
x^2+5x=6-6
x^2+5x=0
x=0