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 the twice-angle formulas, this becomes
$$ -3 \cos(2x) + \sin(2x) = -1 $$
Let $\alpha = \arctan(1/3)$ so $\cos(\alpha) = 3/\sqrt{10}$ and
$\sin(\alpha) = 1/\sqrt{10}$.  The equation can be written as
$$ - \cos(\alpha) \cos(2x) + \sin(\alpha) \sin(2x) = - 1/\sqrt{10}$$
or $$\cos(\alpha + 2 x) = 1/\sqrt{10}$$
