Растолкуйте пожалуйста как решить.2sin(2x+/3)-3cosx=sin(2x)-3

2sin(2x+/3) - 3cos x = sin 2x - 3

2(sin 2xcos /3 + cos 2xsin /3) - 3cos x = sin 2x - 3

2(sin 2x1/2 + cos 2x3/2) - 3cos x = sin 2x - 3

sin 2x + 3cos 2x - 3cos x = sin 2x - 3

3cos 2x - 3cos x = - 3

3(2cosx - 1) - 3cos x = - 3

23cosx - 3 - 3cos x = - 3

23cosx - 3cos x = 0

23cos x(cos x - 3/2) = 0

cos x = 0

x = /2 + n, n Z

cos x - 3/2 = 0

cos x = 3/2

x = (+/-) /6 + 2k, k Z

Ответ: x = /2 + n, x = (+/-) /6 + 2k; n, k Z.