Question

3sinx -4sinxcosx 5cos ^2x =2

Answer 1

3sin^2x-4sinxcosx+5cos^2x=2
3sin^2x-4sinxcosx+5cos^2x = 2cos^2x + 2sin^2
3sin^2x-4sinxcosx+5cos^2x - 2cos^2x - 2sin^2 = 0
sin^2x -4sinxcosx + 3cos^2x = 0
делим на cos^2x
tg^2x - 4tgx + 3 = 0
tgx = y
y^2 - 4y + 3 = 0
y1 = 3
y2 = 1
подставляем
tgx = 3
x = arctg3 + пк, к ∈ z
tgx = 1
x = п/4 + пк, к ∈ z