Sin(2x+П/3)+Cos(2x+П/3)=0
знаем что sin²α+cos²α=1 и sinα+cosα=0 получимsin²α+(-sinα)²=1 => 2sin²α=1 => sin²α=1/2sinα=√2/2
α=π/4+2πn n=0,1,2,3...
α=3π/4+2πn n=0,1,2,3...
и sinα=-√2/2α=5π/4+2πn α=7π/4+2πn n=0,1,2,3...
у sin и cos разные знаки в 2-ом и 4ом четверте
То есть
α=3π/4+2πn 2x+π/3=3π/4+2πn n=0,1,2,3...
α=7π/4+2πn 2x+π/3=7π/4+2πn n=0,1,2,3...
3π/4+2πn и 7π/4+2πn при n=0,1,2,3...
тоже самое 3π/4+πn n=0,1,2,3...
2x+π/3=3π/4+πn n=0,1,2,3...
2x=5π/12+πn n=0,1,2,3...
x=5π/24+πn/2