Cos 2x + 3 sin 2x = 0

Сделаем это уравнение однородным

6sinxcosx+cos^2 (x) - sin^2 (x) - 2sin^2 (x) - 2cos^2 (x) = 0

6sinxcosx-cos^2 (x) - 3sin^2 (x) = 0, разделить на cos^2 (x)

6tgx-1-3tg^2 (x) = 0, tgx=t

6t-1-3t62=0

t1 = (3-K6) / 3, t2 = (3+K6) / 3

1) tgx = (3-K6) / 3, x=arctg ((3-K6) / 3) + pik

2) tgx = (3+K6) / 3, x=arctg ((3+K6) / 3) + pik

Cos 2x + 3Sin 2x = 0 | : Cos 2x≠0

1 + 3 tg 2x = 0

3 tg 2 x = - 1

tg 2x = - 1/3

2x = arc tg (-1/3) + πk, k∈Z

x = - 1/2 arc tg 1/3 + πk, k∈Z