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?

$$\sin2\theta=\cos3^\circ=\sin87^\circ$$
$$\implies2\theta=180^\circ m+(-1)^m87^\circ\text{ where } m \text{ is any integer}$$
$$\implies\theta=90^\circ m+(-1)^m43.5^\circ$$
Check for even $m=2r$(say) and for odd $=2r+1$(say)
Find $m$ such that $\displaystyle\theta\le90^\circ$

Alternatively,
$$\cos3^\circ=\sin2\theta=\cos\left(90^\circ-2\theta\right)=\cos\left(2\theta-90^\circ\right)$$
$$\iff2\theta-90^\circ= 360^\circ n\pm3^\circ\text{ where } n \text{ is any integer}$$
$\displaystyle'+'\implies 2\theta= 360^\circ n+93^\circ\iff\theta=180^\circ n+46.5^\circ$
$\displaystyle'-'\implies 2\theta= 360^\circ n+87^\circ\iff\theta=180^\circ n+43.5^\circ$
Find $n$ such that $\displaystyle\theta\le90^\circ$
