2sin^2x+sinxcosx-3cos^2x=0; |:cos^2x
2tg^2x+tgx-3=0; |:2;
пусть tgx=t
2t^2 + x - 3 = 0
D = b2 - 4ac
D = 1 + 24 = 25 = 5^2
t1,2 = -b ± √D/2a
t1 = -1 + 5/4=1
t2 = -1 - 5/4=-3/2
tgx=-1,5; tgx=1.
x=-arctg(1,5)+pin. n∈Z. x=π/4+pin. n∈Z.
[П/2 : 3П/2]==>>>>>>корни которы входять pi-arctg(3/2) и 5pi/4