Sines everywhere

We see that $cos\big(\frac{3\pi}{10}-\frac{x}{2}\big)=0$ is not the solution, so now with $cos\big(\frac{3\pi}{10}-\frac{x}{2}\big)\neq 0 \Leftrightarrow x \neq \frac{-2\pi}{5}+k2\pi (k\in\mathbb{Z})$ , we multiply both side with it and get $sin\bigg(\frac{3\pi}{5}-x\bigg)=sin\bigg(\frac{\pi}{10}+\frac{3x}{2}\bigg)cos\bigg(\frac{3\pi}{10}-\frac{x}{2}\bigg)$ $\Leftrightarrow sin\bigg(\frac{3\pi}{5}-x\bigg)=\frac{1}{2}\bigg[sin\bigg(\frac{2\pi}{5}+x\bigg)+sin\bigg(2x-\frac{\pi}{5}\bigg)\bigg]$ Set $x+\frac{2\pi}{5}=t$ and we get $sin(\pi-t)=\frac{1}{2}[sin(t)+sin(2t-\pi)]$ The equation above gives us $sin(t)=0$ and $cos(t)=\frac{-1}{2}$ . Solve them and we have $x=\big\{\frac{-2\pi}{5}+k\pi; \frac{4\pi}{15}+k2\pi; \frac{-16\pi}{15}+k2\pi | k\in\mathbb{Z}\big\}$ . Due to the condition, we'll have $x=\bigg\{\frac{3\pi}{5}+k2\pi;\frac{4\pi}{15}+k2\pi; \frac{-16\pi}{15}+k2\pi | k\in\mathbb{Z}\bigg\}$ Since $x\in [-\pi;\pi]$ , we'll get 3 roots $x=\big\{\frac{3\pi}{5};\frac{4\pi}{15}; \frac{14\pi}{15}\big\}$