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?
 A: Another way to solve the problem: note that
$$\sin(2\theta)=\cos(3)=\sin(90-3)$$
But we know $\theta\leq90°$, so $2\theta\leq180°$. Then, two solutions are possible:
$$2\theta=\left\{\begin{array}{ccc}90-3=87°&\Rightarrow&\theta_1=43.5°\\90+3=93°&\Rightarrow&\theta_2=46.5°\end{array}\right.$$
A: $$\sin{( 2 \theta)} = \cos{(3^{\circ})} \Rightarrow \sin{(2 \theta)}=\sin{ \left (90^{\circ} \pm 3^{\circ} \right )} \Rightarrow 2 \theta=90^{\circ} \pm 3^{\circ}+360^{\circ}k, k \in \mathbb{Z}  \\ \Rightarrow \theta=\left ( 45\pm\frac{3}{2}+180k \right )^{\circ}, k \in \mathbb{Z}$$
$$\Rightarrow \theta=\left ( \frac{93}{2}+180k \right )^{\circ} \text{ or } \theta=\left ( \frac{87}{2}+180k \right )^{\circ}, \ \ \ k\in \mathbb{Z}$$
Then you have to find a $\theta$ such that $\theta \leq 90^{\circ}$.
A: Assuming you mean $3^{\circ}$, here are the steps
$$ \sin 2\theta=\cos 3^{\circ} $$
$$ \arcsin(\sin 2\theta)=\arcsin(\cos 3^{\circ}) $$
$$ 2\theta=\arcsin(\cos 3^{\circ}) $$
$$ \theta=\frac{1}{2}\arcsin(\cos 3^{\circ})=\frac{87^{\circ}}{2}=43.5^{\circ} $$
A: 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.)
A: $$\sin2\theta=\cos3^\circ=\sin87^\circ$$
$$\implies2\theta=180^\circ m+(-1)^m87^\circ\text{ where } m \text{ is any integer}$$
$$\implies\theta=90^\circ m+(-1)^m43.5^\circ$$
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 $\displaystyle\theta\le90^\circ$
