Solve for theta , $0<\theta<90^\circ$ 
Solve $$\cos60=\frac{\cos⁡(2θ)+\frac32\sin⁡(2θ))}{\sqrt{4-\sin^2 (2θ)}},$$  where $0<θ<90^\circ $

Question: Is the following solution correct?
Solution attempt:
$$\sqrt{4-\sin^2 (2θ)}=2\cos⁡(2θ)+3\sin⁡(2θ)$$
$$4-\sin^2 (2θ)=(2\cos⁡(2θ)+3\sin⁡(2θ))^2$$
$$4-\sin^2 (2θ)=4\cos^2 (2θ)+12 \sin⁡(2θ)  \cos⁡(2θ)+9\sin^2 (2θ)$$
$$4=4\cos^2 (2θ)+12\sin⁡(2θ)  \cos⁡(2θ)+10\sin^2 (2θ)$$
$$4=4(1-\sin^2 (2θ) +12 \sin⁡(2θ)  \cos⁡(2θ)+10\sin^2 (2θ)$$
$$6\sin^2 (2θ)+12 \sin⁡(2θ)  \cos⁡(2θ)=0$$
$$\sin⁡(2θ) (\sin⁡(2θ)+2 \cos⁡(2θ) )=0$$
$$\sin⁡(2θ)=0 \text{ or } \sin⁡(2θ)+2 \cos⁡(2θ)=0$$
$$θ=0 \ \text{or} \ \tan⁡(2θ)+2=0$$
$θ=0$ or $\tan^{-1} (-2)=-63.4$
$θ=0$ or $2θ=-63.4, \theta=-31.7$
 A: When you arrive to $$\sin⁡(2θ)=0 \text{ or } \sin⁡(2θ)+2 \cos⁡(2θ)=0$$ you should conclude as follows:
Since $\theta \in (0,90^\circ)$, then $2\theta \in (0,180^\circ)$, so $\sin(2\theta)\ne 0$.
From the other equation, $\tan(2\theta) = -2$, so $2\theta = \arctan(-2) + 180^\circ k$ for some integer $k$. Then $\theta = \frac{1}{2}(\arctan(-2) + 180^\circ k)$, and the only value of $k$ that makes $\theta \in (0,90^\circ)$ is $k=1$, so $\theta = \frac{1}{2}(180^\circ-\arctan(2)) = 58.2825\ldots^\circ$
A: Noting that $\mathbf{0^\circ<2\theta<180^\circ}$ and continuing from

$$\sin⁡(2θ)=0 \text{ or } \sin⁡(2θ)+2 \cos⁡(2θ)=0$$

$2\theta=\pi\;$ or $\;\big[\tan⁡(2θ){=}-2\;$ or $\;\cos2\theta=0\big]$
$\displaystyle\theta=\frac\pi2\;$ or $\;\big[2\theta=180^\circ{-}\arctan⁡2\;$ or $\displaystyle\;\frac\pi2\big]$
$\theta\displaystyle=\frac\pi4\;$ or $\displaystyle\;90^\circ{-}\frac12\arctan2\;$ or $\displaystyle\;\frac\pi2$
Of these three candidate solutions, only the middle one, $$58.3^\circ,$$ actually satisfies the given equation.
