Помогите,умоляю! 6 sin^2x+sinx cos x- cos^2x=0

6sin^2x+sinxcosx-cos^2=0 | : cos^2x не равен 0
6tg^2x+tgx-1=0
tgx=t
6t^2+t-1=0
D=1^2-4*6*(-1)=1+24=25
t1;2= $\frac{-1+- \sqrt{25} }{6*2}$
t1= $\frac{-1+5}{12} = \frac{4}{12} = \frac{1}{3}$
t2= $\frac{-1-5}{12} = -\frac{6}{12} =- \frac{1}{2}$

tgx= $\frac{1}{3}$
tgx= $-\frac{1}{2}$

x=arctg( $\frac{1}{3}$ )+ $\pi n$ , n ∈ Z
x=-arctg( $\frac{1}{2}$ )+ $\pi n$ , n ∈ Z