Решить уравнение 2sin^2x-3sinxcosx+4cos^2x=4

2sin^2 (x) + 2cos^2 (x) - 3sinxcosx+2cos^2 (x) = 4

2-3sinxcosx+2cos^2 (x) = 4

2 (1-sin^2 (x)) - 3sinxcosx+2=4

4-2sin^2 (x) - 3sinxcosx=4

-2sin^2 (x) - 3sinxcosx=0

2sin^2 (x) + 3sinxcosx=0

sinx (2sinx+3cosx) = 0

sinx=0

x=pi*n

2sinx+3cosx=0

tgx=-3/2

x=-arctg3/2+pi*n