Trigonometric equation $\sin(60^\circ-2X)\sin(5X)=\sin(8X)\sin(3X)$ A trigonometric equation is to be solved, the solution ($X=10^\circ$) is very clear but I need a proper method
$$\sin(60^\circ-2X)\sin(5X)=\sin(8X)\sin(3X).$$

Using Case $3A$ of the other answer,
let $\sin3x=\dfrac12\implies 3x=180^\circ n+(-1)^n30^\circ\iff x=60^\circ n+(-1)^n10^\circ\ \ \ \ (A)$ where $n$ is any integer
So we have $$\sin8x=2\sin(60^\circ-2x)\sin5x=\cos(60^\circ-7x)-\cos(60^\circ+3x)$$
$(i)$ Set $\cos(60^\circ+3x)=0\implies60^\circ+3x=(2m+1)90^\circ\iff x=60^\circ m+10^\circ \ \ \ \ (B)$ where $m$ is any integer
and we need $\cos(60^\circ-7x)=\sin8x=\cos(90^\circ-8x)$
$\implies60^\circ-7x=360^\circ r\pm(90^\circ-8x)$  where $r$ is any integer
Taking $'+'$ sign, $60^\circ-7x=360^\circ r+(90^\circ-8x)\iff x=360^\circ r+30^\circ\ \ \ \ (C1)$
Taking $'-'$ sign, $60^\circ-7x=360^\circ r-(90^\circ-8x)\iff -15x=360^\circ r-150^\circ\iff x=-24^\circ r+10^\circ\ \ \ \ (C2)$
Combining $A,B,C1;$ we don't have any solution!
Combining $A,B,C2; x=120^\circ r+10^\circ$
$(ii)$ Try setting $\cos(60^\circ-7x)=0$
