Ответ:
x1 = -pi/4 + pi*n; n ∈ Z; x2 = pi/6 + 2pi*k; k ∈ Z; x3 = 5pi/6 + 2pi*k; k ∈ Z
Объяснение:
2(sin x + cos x) = cos x/sin x + sin x/sin x
2(sin x + cos x) = (cos x + sin x)/sin x
2(sin x + cos x) - (cos x + sin x)/sin x = 0
(sin x + cos x)(2 - 1/sin x) = 0
1) sin x + cos x = 0
Делим все на cos x ≠ 0
sin x/cos x + 1 = 0
tg x = -1
x1 = -pi/4 + pi*n; n ∈ Z
2) 2 = 1/sin x
sin x = 1/2
x2 = pi/6 + 2pi*k; k ∈ Z
x3 = 5pi/6 + 2pi*k; k ∈ Z