Решение
cos3x+cos2x=sin5x
2cos(3x + 2x)/2* cos(3x - 2x)/2 = sin5x
2cos2,5x * cos0,5x - sin2*2,5x = 0
2*cos2,5x * cos0,5x - 2*sin2,5x * cos2,5x = 0
2cos2,5x * (cos0,5x - sin2,5x) = 0
1) cos2,5x = 0
2,5x = π+ πk, k ∈ Z
x₁ = 2π/5 + 2πk/5, k ∈ Z
2) cos0,5x - sin2,5x = 0
cos0,5x - cos(π/2 - 2,5x) = 0
- 2sin(0,5x + 2,5x)/2 * sin(0,5x - 2,5x)/2 = 0
3) sin1,5x = 0
1,5x = πn, n ∈ Z
x₂ = 2πn/3, n ∈ Z
4) sinx = 0
x₃ = πm, m ∈ Z