Solve the equation $\sin 3 x+3 \cos x=2 \sin 2 x(\sin x+\cos x)$.
Solution. In the equation $\sin 3 x+3 \cos x=2 \sin 2 x \cdot(\sin x+\cos x)$, we first apply the addition theorem as follows:

$$
\begin{aligned}
\sin (2 x+x)+3 \cos x & =2 \sin 2 x \cdot \sin x+2 \sin 2 x \cdot \cos x \\
\cos 2 x \cdot \sin x+\sin 2 x \cdot \cos x+3 \cos x & =2 \sin 2 x \cdot \sin x+2 \sin 2 x \cdot \cos x
\end{aligned}
$$

Using the double-angle formulas for sine and cosine, we arrive at the following equation:

$$
\cos ^{2} x \cdot \sin x-\sin ^{3} x+3 \cos x=4 \sin ^{2} x \cdot \cos x+2 \sin x \cdot \cos ^{2} x
$$

After rearranging to zero and possible simplifications:

$$
\begin{array}{r}
4 \sin ^{2} x \cdot \cos x+\sin x \cdot\left(\cos ^{2} x+\sin ^{2} x\right)-3 \cos x=0 \\
4 \sin ^{2} x \cdot \cos x+\sin x-3 \cos x=0
\end{array}
$$

It is easily verifiable that $\cos x=0$ does not lead to a solution, so we can divide both sides of the equation by $\cos x$:

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

Then, using the identity $\sin ^{2} x=\frac{\operatorname{tg}^{2} x}{\operatorname{tg}^{2} x+1}$, we get:

$$
4 \cdot \frac{\operatorname{tg}^{2} x}{\operatorname{tg}^{2} x+1}+\operatorname{tg} x-3=0
$$

Rearranging the equation, we arrive at a cubic equation in $\operatorname{tg} x$, one of whose solutions is apparently 1:

$$
(\operatorname{tg} x-1)(\operatorname{tg} x+2 \operatorname{tg} x+3)=0
$$

The discriminant of the quadratic expression is negative, so the only solution to the above equation is:

$$
\operatorname{tg} x=1, \quad x=\frac{\pi}{4}+k \pi, \quad \text { where } \quad k \in \mathbb{Z}
$$

Solution by Tóth 156 Sándor (Csongrád, Batsányi J. Gymn., 12th grade)