Problems from Solving trigonometric equations

Development:

Let's remember the formula that transforms the sum into a product: $$\sin (A)+\sin (B)=2\cdot \sin ({\displaystyle \frac{A+B}{2}})\cdot \cos ({\displaystyle \frac{A-B}{2}})$$

Applying it in our case, it gives: $$\sin (2x+{\displaystyle \frac{\pi }{3}})+\sin (x+{\displaystyle \frac{\pi }{6}})=2\cdot \sin ({\displaystyle \frac{2x+{\displaystyle \frac{\pi }{3}}+x+{\displaystyle \frac{\pi }{6}}}{2}})\cdot \cos ({\displaystyle \frac{2x+{\displaystyle \frac{\pi }{3}}-x-{\displaystyle \frac{\pi }{6}}}{2}})=$$ $$=2\cdot \sin ({\displaystyle \frac{3x+{\displaystyle \frac{\pi }{2}}}{2}})\cdot \cos ({\displaystyle \frac{x+{\displaystyle \frac{\pi }{6}}}{2}})=2\cdot \sin ({\displaystyle \frac{3x}{2}}+{\displaystyle \frac{\pi }{4}})\cdot \cos ({\displaystyle \frac{x}{2}}+{\displaystyle \frac{\pi }{12}})=0$$

So if the product of two factors is zero, necessarily one of the two is zero. Therefore, we distinguish:

case (a): $$\sin ({\displaystyle \frac{3x}{2}}+{\displaystyle \frac{\pi }{4}})=0\Rightarrow {\displaystyle \frac{3x}{2}}+{\displaystyle \frac{\pi }{4}}=\{\begin{array}{c}2\pi \cdot k\\ \pi +2\pi \cdot k\end{array}k\in \mathbb{Z}=k\pi ,k\in \mathbb{Z}\Rightarrow$$ $$\Rightarrow {\displaystyle \frac{3x}{2}}=k\pi -{\displaystyle \frac{\pi }{4}}\Rightarrow x=-{\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{2}{3}}\pi \cdot k$$

and case (b): $$\cos ({\displaystyle \frac{x}{2}}+{\displaystyle \frac{\pi }{12}})=0\Rightarrow {\displaystyle \frac{x}{2}}+{\displaystyle \frac{\pi }{12}}=\{\begin{array}{c}{\displaystyle \frac{\pi }{2}}+2k\cdot \pi \\ -{\displaystyle \frac{\pi }{2}}+2k\cdot \pi \end{array}k\in \mathbb{Z}={\displaystyle \frac{\pi }{2}}+k\pi ,k\in \mathbb{Z}\Rightarrow$$ $$\Rightarrow {\displaystyle \frac{x}{2}}={\displaystyle \frac{\pi }{2}}+k\cdot \pi -{\displaystyle \frac{\pi }{12}}\Rightarrow x={\displaystyle \frac{5\pi }{6}}+2k\cdot \pi$$

Solution:

$x=\{\begin{array}{c}-{\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{2}{3}}k\cdot \pi \\ {\displaystyle \frac{5\pi }{6}}+2k\cdot \pi \end{array},k\in \mathbb{Z}$

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