Solve the following trigonometric equation:
$$\sin \Big( 2x+\dfrac{\pi}{3}\Big)+\sin \Big(x+\dfrac{\pi}{6} \Big)=0$$
Development:
Let's remember the formula that transforms the sum into a product: $$$\sin(A)+\sin(B)=2\cdot\sin\Big(\dfrac{A+B}{2}\Big)\cdot\cos\Big( \dfrac{A-B}{2}\Big)$$$
Applying it in our case, it gives: $$$\sin(2x+\dfrac{\pi}{3})+\sin(x+\dfrac{\pi}{6})=2\cdot\sin\Big(\dfrac{2x+\dfrac{\pi}{3}+x+\dfrac{\pi}{6}}{2}\Big)\cdot\cos\Big(\dfrac{2x+\dfrac{\pi}{3}-x-\dfrac{\pi}{6}}{2}\Big)=$$$ $$$=2\cdot\sin\Big(\dfrac{3x+\dfrac{\pi}{2}}{2}\Big)\cdot\cos\Big(\dfrac{x+\dfrac{\pi}{6}}{2}\Big)=2\cdot\sin\Big(\dfrac{3x}{2}+\dfrac{\pi}{4}\Big)\cdot\cos\Big(\dfrac{x}{2}+\dfrac{\pi}{12}\Big)=0$$$
So if the product of two factors is zero, necessarily one of the two is zero. Therefore, we distinguish:
case (a): $$$\sin\Big(\dfrac{3x}{2}+\dfrac{\pi}{4}\Big)=0 \Rightarrow \dfrac{3x}{2}+\dfrac{\pi}{4}= \left\{\begin{array}{c} 2\pi\cdot k \\ \pi+2\pi\cdot k \end{array}\right. k \in \mathbb{Z}=k\pi, \ k\in\mathbb{Z} \Rightarrow $$$ $$$\Rightarrow \dfrac{3x}{2}=k\pi-\dfrac{\pi}{4} \Rightarrow x=-\dfrac{\pi}{6}+\dfrac{2}{3}\pi\cdot k$$$
and case (b): $$$\cos\Big(\dfrac{x}{2}+\dfrac{\pi}{12}\Big)=0 \Rightarrow \dfrac{x}{2}+\dfrac{\pi}{12}= \left\{\begin{array}{c} \dfrac{\pi}{2}+2k\cdot\pi \\ -\dfrac{\pi}{2}+2k\cdot\pi \end{array}\right. k \in \mathbb{Z}=\dfrac{\pi}{2}+k\pi, \ k\in\mathbb{Z} \Rightarrow $$$ $$$\Rightarrow \dfrac{x}{2}=\dfrac{\pi}{2}+k\cdot\pi-\dfrac{\pi}{12} \Rightarrow x=\dfrac{5\pi}{6}+2k\cdot\pi$$$
Solution:
$$x=\left\{\begin{array}{c} -\dfrac{\pi}{6}+\dfrac{2}{3}k\cdot\pi \\ \dfrac{5\pi}{6}+2k\cdot\pi \end{array}\right. ,k \in \mathbb{Z}$$