Sin(5/3x+p/4)=-корень из 3/2

Решите задачу:

$sin(\frac{5x}3+\frac{\pi}4)=-\frac{\sqrt3}2$
$\left[\begin{array}{ccc}\frac{5x}3+\frac{\pi}4=-\frac{\pi}3+2\pi n;n\in Z\\\frac{5x}3+\frac{\pi}4=-\frac{2\pi}3+2\pi n;n\in Z\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}\frac{5x}3=-\frac{\pi}3-\frac{\pi}4+2\pi n;n\in Z\\\frac{5x}3=-\frac{2\pi}3-\frac{\pi}4+2\pi n;n\in Z\end{array}\right$

$=\ \textgreater \ \left[\begin{array}{ccc}\frac{5x}3=-\frac{4\pi}{12}-\frac{3\pi}{12}+2\pi n;n\in Z\\\frac{5x}3=-\frac{8\pi}{12}-\frac{3\pi}{12}+2\pi n;n\in Z\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}\frac{5x}3=-\frac{7\pi}{12}+2\pi n;n\in Z\\\frac{5x}3=-\frac{11\pi}{12}+2\pi n;n\in Z\end{array}\right$

$=\ \textgreater \ \left[\begin{array}{ccc}5x=-\frac{7\pi}{4}+6\pi n;n\in Z\\\ 5x=-\frac{11\pi}{4}+6\pi n;n\in Z\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}x=-\frac{7\pi}{20}+\frac{6\pi n}{5};n\in Z\\\ x=-\frac{11\pi}{20}+\frac{6\pi n}{5};n\in Z\end{array}\right$