Solve the following equation:

$$
3 \sin ^{2} x-4 \cos ^{2} x=\frac{\sin 2 x}{2} .
$$
Let's replace $\sin 2 x$ with $2 \sin x \cos x$ and divide the equation by $\cos ^{2} x$. (We can do this because $\cos x=0$ is not possible. Indeed, if $\cos x=0$, then $\sin ^{2} x=1$ and thus the right side of our equation is 0, while the left side is not 0.)

$$
3 \operatorname{tg}^{2} x-4=\operatorname{tg} x
$$

that is,

$$
3 \operatorname{tg}^{2} x-\operatorname{tg} x-4=0
$$

from which

$$
\operatorname{tg} x_{1}=-1, \quad \operatorname{tg} x_{2}=\frac{4}{3}
$$

and thus

$$
x_{1}=135^{\circ} \pm k \cdot 180^{\circ}, \quad x_{2}=53^{\circ} 7^{\prime} \pm k \cdot 180^{\circ}(k=0,1,2 \ldots)
$$

By substitution, we have verified that these roots indeed satisfy the given equation.