Помогите решить уравнение:

Sin (2x+pi/3) + cos (2x+pi/3) = 0.

Sin (2x+π/3) + sin (π/6-2x) = 0

2sinπ/2cos (2x+π/12) = 0

cos (2x+π/12) = 0

2x+π/12=π/2+πn

2x=5π/12+πn

x=5π/24+πn/2

Воспользуемся формулой sin x + cos x = √2cos (45-x).

Тогда sin (2x+pi/3) + cos (2x+pi/3) = √2cos (π/4-2x-π/3) =

= √2cos (-π/12-2x) = - √2cos (π/12+2x) = 0.

После сокращения на - √2 получим:

cos (π/12+2x) = 0 π/12+2x = 2kπ+-π/2.

2x₁ = 2kπ+π/2-π/12 = 2kπ+5π/12. x₁ = kπ+5π/24,

2x₂ = 2kπ-π/2-π/12 = 2kπ-7π/12. x₂ = kπ-7π/24.