Question

3 sin квадрат x - 4 sin x cos x +5 cos квадрат x = 2
Помогите решить

Answer 1

3sin^2x - 4sinxcosx + 5cos^2x = 2*1
3sin^2x - 4sinxcosx + 5cos^2x = 2*(sin^2x+cos^2x)
3sin^2x - 4sinxcosx + 5cos^2x = 2sin^2x + 2cos^2x
3sin^2x - 2sin^2x - 4sinx*cosx + 5cos^2x - 2cos^2x = 0   
sin^2x - 4sinx*cosx + 3cos^2x = 0   // : cos^2x ≠ 0 
tg^2x - 4tgx + 3 = 0 
(tgx - 3) (tgx - 1) = 0 

tgx = 3 
x = arctg(3) + pik, k ∈Z

tgx = 1 
x = pi/4 + pik, k ∈Z