Решите уравнение:

3sin^2 2x-3sin2x=sin^2 2x+2

3sin^2 х-4sinxcosx+5cos^2x-2=0

3sin^2 х-4sinxcosx+5cos^2x-2• (sin^2x+cos^2x) = 0

3sin^2 х-4sinxcosx+5cos^2x-2sin^2x-2cos^2x=0

Sin^2x-4sinxcosx+3cos^x=0. (:cos^2x)

tg^2x-4tgx+3=0

tgx=t

t^2-4t+3=0

D=16-12=4

t1 = (4+2) : 2=3

t2 = (4-2) : 2=1

tgx=1

X=arctg1+Пк; к€Z

X=П/4+Пк; к€Z

tgx=3

X=arctg3+Пк; к€Z