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.