Question

2sinxcosx+3cos^2x-sin^2x=0

Answer 1

2sinxcosx+3cos^2x-sin^2=0 | :cos^2x
2tgx+3-tg^2x=0
tgx=t
-t^2+2t+3=0 | : -1
t^2-2t-3=0
D=4+12=16
t 1,2= 2+-4/2
t1=3; t2=-1
tgx=3
x=arctg3+pi*n; n ∈ Z
tgx=-1
x=-pi/4 + pi*n; n ∈ Z