решить уравнение cos5x-cosx=sin2x

cos(5x)-cos(x)=sin(2x)

-ssin((5x+x)2)*sin((5x-x)/2)=sin(2x)

-2sin(3x)*sin(2x)=sin(2x)

2sin(3x)*sin(2x)-sin(2x)=0

sin(2x)*(2sin(3x)-1)=0

a) sin(2x)=0 => 2x=pi*n => x=pi*n/2

б) 2sin(3x)+1=0 => 2sin(3x)=-1 => sin(3x)=-1/2 => 3x=(-1)^n*arcsin(-1/2)+pi*n => 3

    => x=(-1)^n*(-pi/6) +pi*n => x=(-1)^n*(-pi/18) +pi*n/3

-2 * sin 3x * sin 2x = sin 2x

2 * sin 3x * sin 2x + sin 2x = 0

sin 2x * (2 * sin 3x + 1) = 0

1) sin 2x = 0             2)  sin 3x = -1/2

    2x = π * m                3x = (-1)^n * (-π/6) + π * n

    x = π * m / 2             x = (-1)^(n + 1) * π/18 + π * n / 3