Решение
2sin^2x-3sinxcosx+4cos^2x=4
(2sin²x + 2sin²x) - 2sin²x + 4сos²x - 3sinxcosx = 4
4(sin²x + cos²x) - 2sin²x - 3sinxcosx = 4
4 - 2sin²x - 3sinxcosx = 4
- 2sin²x - 3sinxcosx = 0
- sinx(2sinx + 3cosx) = 0
1) sinx = 0
x₁ = πk, k∈Z
2) 2sinx + 3cosx = 0 делим на cosx≠ 0
2tgx + 3 = 0
tgx = - 3/2
x₂= arctg(-3/2) + πn, n∈Z