Объясните пожалуйста как решить.
2sin(2x+π/3)-3cosx=sin(2x)-√3

2sin(2x+π/3) - 3cos x = sin 2x - √3

2(sin 2x·cos π/3 + cos 2x·sin π/3) - 3cos x = sin 2x - √3

2(sin 2x·1/2 + cos 2x·√3/2) - 3cos x = sin 2x - √3

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

√3cos 2x - 3cos x = - √3

√3(2cos²x - 1) - 3cos x = - √3

2√3cos²x - √3 - 3cos x = - √3

2√3cos²x - 3cos x = 0

2√3cos x·(cos x - √3/2) = 0

cos x = 0

x = π/2 + πn, n ∈ Z

cos x - √3/2 = 0

cos x = √3/2

x = (+/-) π/6 + 2πk, k ∈ Z

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