<s>[INST] Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $ 
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
  $$ 0 \le x \le  360^{\circ} $$


My attempt: 
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2}  - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5  = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
 [/INST] Set $\sin(2x) = u$
$7\cos(2x) - \sin(2x)  + 5 = 0$
$\implies 7\sqrt{1-u^2} = u-5$
$\implies 49 - 49u^2 = u^2 - 10u + 25$
$\implies 50u^2 - 10u -24 = 0$
$\implies u_1 = 0.8, u_2 = -0.6$
$ \sin(2x) = 0.8$ or $\sin(2x) = -0.6$
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