Решить уравнение cosx+cos9x=cos5x+cos7x

(соs9x-cos5x) - (cos7x-cosx) = 0

-2sin7xsin2x+2sin3xsin4x=0

4sin3xsin2xcos2x-2sin7xsin2x=0

2sin2x (2sin3xcos2x-sin7x) = 0

sin2x=0⇒2x=πn, n∈Z⇒x=πn/2, n∈Z

2sin3xcos2x-sin7x=0

2*1/2 * (sinx+sin5x) - sin7x=0

sinx+sin5x-sin7x=0

sinx-2sinxcos6x=0

sinx (1-2cos6x) = 0

sinx=0⇒x=πk, k∈Z

cos6x=1/2⇒6x=+-π/3+2πm, m∈Z⇒x=+-π/18+πm/3, m∈Z

Ответ x=πn/2, n∈Z; x=+-π/18+πm/3, m∈Z