cosx^2-sinx^2=0
cos2x=0
2x=π/2+πn,n∈z
x=π/4+πn/2,n∈z.
ответ: x=π/4+πn/2,n∈z.
Вот!)
cos^2(x)=sin^2(x)
cos^2(x)-sin^2(x)=0
cos(2x)=0
2x=π/2+πk, k∈Z
x=π/4+πk/2, k∈Z
