Solve the following equation:

$$
\sin \left(x+15^{\circ}\right)+\sin \left(x+45^{\circ}\right)+\sin \left(x+75^{\circ}\right)=\sin 15^{\circ}+\sin 45^{\circ}+\sin 75^{\circ} .
$$
Solve the following equation:

$$
\sin \left(x+15^{\circ}\right)+\sin \left(x+45^{\circ}\right)+\sin \left(x+75^{\circ}\right)=\sin 15^{\circ}+\sin 45^{\circ}+\sin 75^{\circ} .
$$

Solution. We apply the identity $\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$ on the left side of equation (1) with the choice $\alpha=x+75^{\circ}, \beta=x+15^{\circ}$:

$$
\sin \left(x+75^{\circ}\right)+\sin \left(x+15^{\circ}\right)=2 \sin \left(x+45^{\circ}\right) \cdot \cos 30^{\circ}
$$

We apply the identity again on the right side of equation (1) with the choice $\alpha=75^{\circ}$ and $\beta=15^{\circ}$,

$$
\sin 75^{\circ}+\sin 15^{\circ}=2 \sin 45^{\circ} \cdot \cos 30^{\circ}
$$

Thus, (1) can be written as:

$$
2 \sin \left(x+45^{\circ}\right) \cos 30^{\circ}+\sin \left(x+45^{\circ}\right)=2 \sin 45^{\circ} \cdot \cos 30^{\circ}+\sin 45^{\circ} .
$$

Factoring out $\sin \left(x+45^{\circ}\right)$ on the left and $\sin 45^{\circ}$ on the right:

$$
\sin \left(x+45^{\circ}\right)\left[2 \cos 30^{\circ}+1\right]=\sin 45^{\circ}\left(2 \cos 30^{\circ}+1\right)
$$

Since $2 \cos 30^{\circ}+1 \neq 0$, we can simplify by it (we do not lose any roots).

We get that

$$
\sin \left(x+45^{\circ}\right)=\sin 45^{\circ}
$$

which holds if

$$
\begin{aligned}
x+45^{\circ} & =45^{\circ}+k \cdot 360^{\circ}, & & \text { i.e., } x=k \cdot 360^{\circ}, \\
\text { or } x+45^{\circ} & =135^{\circ}+k \cdot 360^{\circ}, & & \text { i.e., } x=90^{\circ}+k \cdot 360^{\circ}, \quad(k \in \mathbf{Z})
\end{aligned}
$$

Based on the work of Mónika Illés (Sátoraljaújhely, Kossuth Gymnasium, 2nd year)