<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] I'm not sure why you're using this complicated method. If you have an equation of the form
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d
$$
you can just observe that this is equivalent to
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d\sin^2x+d\cos^2x
$$
so it becomes
$$
(a-d)\sin^2x+b\sin x\cos x+(c-d)\cos^2x=0
$$
This is Roman83's solution, but I'd like to discuss it in general, then comparing it with your method.
Note that $\cos x=0$ gives a solution if and only if $a-d=0$, in which case you get the equation
$$
\cos x(b\sin x+(c-d)\cos x)=0
$$
which is easy. If $a-d\ne0$, you can divide both sides by $\cos^2x$ and get
$$
(a-d)\tan^2x+b\tan x+(c-d)=0
$$
which is a quadratic in $\tan x$.

Your method leads to the same quadratic, but doing a lot of work. First you observe that
$$
\sin^2x=\frac{1-\cos2x}{2},\qquad
\cos^2x=\frac{1+\cos2x}{2}
$$
so you get, after multiplying by $2$ both sides,
$$
a(1-\cos2x)+2b\sin x\cos x+c(1+\cos2x)=2d
$$
hence
$$
(c-a)\cos2x+b\sin2x+a+c-2d=0
$$
Now the Weierstrass substitution gives, setting $t=\tan x$,
$$
(c-a)\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}+a+c-2d=0
$$
that becomes
$$
(c-a)-(c-a)t^2+2bt+(a+c-2d)+(a+c-2d)t^2=0
$$
and, reordering terms,
$$
2(a-d)t^2+2bt+2(c-d)=0
$$
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