I need to solve the following trigonometric equation:
$$\tan (2x) = 6\cos ^2(x) - 4\sin (x)\cos (x) - 2\sin ^2(x)$$
My attempt:
$$\frac{\sin(2x)}{\cos(2x)} = 3(\cos(2x)+1) - 2\sin(2x) -2(1-\cos^2(x))$$
$$\frac{\sin(2x)}{\cos(2x)} = 3\cos(2x)+3 - 2sin(2x) - 2(1-\frac{\cos(2x)+1}{2})$$
$$\frac{\sin(2x)}{\cos(2x)} = 3\cos(2x)+3-2\sin(2x)-2+\cos(2x)+1$$
$$\frac{\sin(2x)}{\cos(2x)} = 4\cos(2x) +2 - 2\sin(2x)$$
Adding $2\sin(2x)$ to each side and then multiplying by $\cos(2x)$ $$\sin(2x)(1+2\cos(2x)) = 4\cos^2(2x)+2\cos(2x)$$
$$\sin(2x)(1+2\cos(2x)) +1= 4\cos^2(2x)+2\cos(2x)+1$$
$$\sin(2x)(1+2\cos(2x))+1 = (2\cos(2x)+1)^2$$
$$1 = (2\cos(2x)+1)) (2\cos(2x)+1 - \sin(2x))$$
After that I got stuck.
You find correctly that $$ \tan2x=4\cos2x-2\sin2x+2 $$ Multiplying by $\cos2x$ leads to $$ 4\cos^22x-2\sin2x\cos2x+2\cos2x-\sin2x=0 $$ that can be rewritten as $$ 2\cos2x(2\cos2x-\sin2x)+(2\cos2x-\sin2x)=0 $$ so $$ (2\cos2x+1)(2\cos2x-\sin2x)=0 $$ This splits into $$ \cos2x=-\frac{1}{2}\qquad\text{or}\qquad \sin2x=2\cos2x $$ and both are elementary.