If $\sin( 2 \theta) = \cos( 3)$ and $\theta \leq 90°$, find $\theta$ Find $\theta\leq90°$ if 
$$\sin( 2 \theta) = \cos( 3)$$
I know that $\sin 2\theta = 2\sin\theta\cos \theta$, or alternatively, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$.
Can somebody help me?

As you've noted, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$. Cosine of $3$ is in the second quadrant and is quite close to $-1$, and sine of something close to $-1$ is going to be in the fourth quadrant, hence this sine is going to be negative. Dividing by two doesn't change the sign, so the aswer is:
$$\theta = -\dfrac{180\sin^{-1}(\cos 3)}{2\pi}=-\dfrac{90\sin^{-1}(\cos 3)}{\pi}$$
(The factor of $\frac{180}{\pi}$ comes from the fact that we need to convert radians to degrees.)
