Sin (2x+П/3) + Cos (2x+П/3) = 0

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 4 ом 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