Решить уравнение cos5x-cosx=sin2x

-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

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