TITLE: Solve trigonometric equation for x
QUESTION [7 upvotes]: Solve for x: $$2 \sin^2 x + \sin 2x = 1$$
I've tried a few strategies, just applying known, standard, equalities to get to some point of understanding. The latest is along the line of trying to work out al cosinuses to sinueses and arrive at a quadratic equation which I could solve using the abc-formula.
This got me stuck with something I don't know how to deal with. I'm really just looking for a small hint on how to get further from here or on starting a new strategy, so I can (try to) figure the rest out myself. It's really important I understand this thoroughly, there are a few exercises left in this paragraph I have issues with and I want to be able to solve them all by myself. TL;DR.
$$2 \sin^2 x + \sin 2x = 1$$
$$ 2 \sin^2 x + 2 \cos x \sin x = 1$$
Since this isn't my first attempt, I'm just trying to get somewhere...
$$ 2 \sin^2 x + 2 \cos x \sin x = \cos^2x + \sin^2 x$$
$$ \sin^2 x + 2 \cos x \sin x - \cos^2 x = 0$$
The above does look alot like something familiar, so I thought perhaps if I get rid of the cosinus...
$$ \sin^2 x + 2 \sin(\frac{\pi}{2} - x) \sin x - (1 - \sin^2 x) = 0$$
$$ 2 \sin^2 x + 2 \sin(\frac{\pi}{2} - x) \sin x - 1 = 0$$
Now I don't know how to progess any further, what to do with $\sin(\frac{\pi}{2} - x)$, for example. My strategy is probably just wrong. Where to go from here?

REPLY [4 votes]: $2 \sin^2 x + \sin 2x = 1\Rightarrow 2 \sin^2 x + \sin 2x =\sin^2 x+\cos^2 x\Rightarrow \sin 2x=\cos 2x \Rightarrow \tan 2x=1$