<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] $7cos(2x)-sin(2x)+5=0$
$7cos(2x)/\sqrt{50}-sin(2x)/\sqrt{50}=-5/\sqrt{50}$
Let $\alpha=sin^{-1}(7/\sqrt{50})$
Then
$sin(\alpha)cos(2x)+cos(\alpha)sin(2x)=-5/\sqrt{50}$
$sin(2x+\alpha)=-5/\sqrt{50}$
$2x+\alpha=sin^{-1}(-1/\sqrt{2})$
So
$x=(sin^{-1}(-1/\sqrt{2})-\alpha)/2$
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