Question

6sin^2+sinx × cosx - cos^2x=0

Answer 1

разделим на cos²x≠0
6tg²x+tgx-1=0
tgx=a
6a²+a-1=0
D=1+24=25
a1=(-1-5)/12=-1/2⇒tgx=-1/2⇒x=-arctg1/2+πn
a2=(-1+5)/12=1/3⇒tgx=1/3⇒x=arctg1/3+πn