As David Mitra observed in the comments, by dividing through by $\sin\theta$, you lost those solutions in which $\sin\theta = 0$. You also failed to find another solution by failing to consider symmetry.
\begin{align*}
4\sin\theta & = 3\tan\theta\\
4\sin\theta - 3\tan\theta & = 0\\
4\sin\theta - 3\frac{\sin\theta}{\cos\theta} & = 0\\
4\sin\theta\cos\theta - 3\sin\theta & = 0\\
\sin\theta(4\cos\theta - 3) & = 0
\end{align*}
\begin{align*}
\sin\theta & = 0 & 4\cos\theta - 3 & = 0\\
\theta & = n\pi, n \in \mathbb{Z} & 4\cos\theta & = 3\\
& & \cos\theta & = \frac{3}{4}\\
& & \theta & = \pm \arccos\left(\frac{3}{4}\right) + 2m\pi, m \in \mathbb{Z}
\end{align*}
The requirement that $-180^\circ \leq \theta \leq 180^\circ$ is equivalent to $-\pi \leq \theta \leq \pi$. Hence, we must take $n = -1, 0, 1$, which yields
$\theta = -\pi, 0, \pi$ or, equivalently, $\theta = -180^\circ, 0^\circ, 180^\circ$.
Since the function $f: [-1, 1] \to [0, \pi]$ defined by $f(x) = \arccos x$ yields the unique angle $\theta$ in the interval $[0, \pi]$ such that $\cos\theta = x$, there is one angle in the interval $[0, \pi]$ such that $\theta = \arccos\left(\frac{3}{4}\right)$. It is $\approx 0.7227342478$ radians, which converts to $\approx 41.4^\circ$ according to this calculator. However, we want all the solutions in the interval $-180^\circ \leq \theta \leq 180^\circ$. Since the cosine function is even, $\cos(-x) = \cos x$. Thus, $\theta = -\arccos\left(\frac{3}{4}\right) \approx -41.4^\circ$ is also a solution. There are no other solutions in the interval since $-180^\circ \leq \theta \leq 180^\circ \implies m = 0$.
