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.
 A: $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot 1 $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 \cdot (\sin^2(x)+\cos^2(x)) $$
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5\sin^2(x)+5\cos^2(x) $$
$$ \sin^2(x)+\sin(x)\cos(x)-6\cos^2(x)=0$$
$$\tan^2(x)+\tan(x)-6=0$$
$\tan(x)=2$ or $\tan (x)=-3$
$x=\arctan2+\pi n, n \in \mathbb Z$
$x=-\arctan3+\pi k, k \in \mathbb Z$
A: It was a good work arriving to $$7\cos(2x) - \sin(2x) + 5 = 0$$ But the substitution to be used was just $t=\tan(\frac{2x}2)=\tan(x)$ which leads to the quadratic $$t^2+t-6=0$$ leading to $$(t-2)(t+3)=0$$
A: HINT
Perhaps the other methods are easier but to continue where you left off, 
Realize that $$\sin 2x=7\cos 2x+5$$
Use $$\sin^2 2x+\cos^2 2x=1$$
To make your last equation into a quadratic for $\cos 2x$.
A: hint: Divide both sides by $\cos^2 x$ and get an quadratic equation in $\tan x$
A: HINT : 
Dividing the both sides by $\cos^2\theta\ (\not=0)$ gives
$$6\tan^2(x)+\tan(x)-1=\frac{5}{\cos^2(x)}=5(1+\tan^2(x))$$
A: 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 $\sin(2x) = -0.6$
A: 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,
$$
2(a-d)t^2+2bt+2(c-d)=0
$$
A: $$ 6 s^2 + s c -c^2 =5 $$
Divide by c^2 . $ t=s/c,$
$$ 6 t^2 + t - 1 = 5/c^2  = 5 ( 1 +t^2) \rightarrow ;\; t^2 + t -6 =0;\; (t-2)(t+3) =0$$
$$ t= \tan x = 2,-3 $$
A: $7cos(2x)-sin(2x)+5=0$
$7cos(2x)/\sqrt{50}-sin(2x)/\sqrt{50}=-5/\sqrt{50}$
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
$x=(sin^{-1}(-1/\sqrt{2})-\alpha)/2$
