<s>[INST] find the solutions to the equation $4\sin^2\theta + 1 = 6\sin\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ Find the solutions to the following equation for $0^\circ \leq θ < 360^\circ$:
$$4\sin^2 θ + 1 = 6\sin θ$$
My work:
$$4\sin^2\theta - 6\sin\theta + 1 = 0$$
Factor
$$\sin\theta= \frac{1}{4}(3+ \sqrt{5}) \qquad \text{and} \qquad \sin\theta = \frac{1}{4}(3- \sqrt{5})$$
$$\theta = \arcsin\left[\frac{1}{4}(3+ \sqrt{5})\right] \qquad \text{and} \qquad \theta = \arcsin\left[\frac{1}{4}(3- \sqrt{5})\right]$$
When I take the inverse sin they both come out >1 (no solution), what did I do wrong?

\begin{align*}
2\cos^2\theta - 2\cos\theta - 1 & = 0\\
\cos\theta & = \frac{2 \pm \sqrt{4+8}}{4} && \text{quadratic formula}\\
\cos\theta & = \frac{1 \pm \sqrt{3}}{2}\\
\theta & = \arccos\left[\frac{1-\sqrt{3}}{2}\right]\\
\theta & = 111.5^\circ
\end{align*}
Is this correct?
 [/INST] The domain of arcsin is $[-1,1]$, the co-domain is $[-\pi/2, \pi/2]$
So you have only the solution $a = arcsin((3-\sqrt5)/4)$. That's because you cannot take arcsin from $(3+\sqrt5)/4$, as it's not in the domain. 
Then ... given that your $sin$ value $(3-\sqrt5)/4$ is in $(0, 1)$, you conclude that $0 < a < \pi/2$. From that follows that $\pi - a$ is also a solution. And obviously you have no more solutions in $[0, 2\pi)$. So you have 2 solutions here: $a$ and $\pi-a$. 
</s>