Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$ I tried 
$$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$
Simplifying, 
$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$
Then I'm stuck!

Using that $$-2 = -2\sin^2(x) - 2\cos^2(x)$$ from $$6\sin^2(x) + 2\cos(x)\sin(x) - 2 = 0$$ we get $$4\sin^2(x) + 2\cos(x)\sin(x) - 2\cos^2(x) = 0$$ and dividing by $\cos^2(x)$ we have $$4\tan^2(x) + 2\tan(x) - 2 = 0$$
Can you continue from here? Note that when dividing by $\cos^2(x)$ we assume that it is not $0$. What if it is $0$?
