Solving $\sin x=2\sin(2\pi/3-x)$ How can I solve the equation: $$\sin x=2\sin\left(\frac{2\pi}{3}-x\right)$$
Without using the formula:
$$\sin(a-b)=\sin a \cos b-\sin b \cos a$$?
Thanks.

Consider the next picture 

By Sine Law the equality $\frac{\sin x}{2} = \frac{\sin\left(2\pi/3-x\right)}{1}$ holds, then by using the Cosine Law we have
\begin{align*}
c^2&=5-4\cos\frac{\pi}{3}\\
&=3\\
c&=\sqrt{3}
\end{align*}
Now, from Sine Law, it follows.
\begin{align*}
\sin x&=\frac{2\sin \pi/3}{\sqrt{3}}\\
\sin x&=1\\
\implies x&=\frac{\pi}{2}
\end{align*}
