Prove the identity $\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)=\sin x$. Hence solve the equation $\sin\left(\frac{\pi}{3}+2x\right)-\sin\left(\frac{\pi}{3}-2x\right)-2=4\sin 2x$ for $0 \le x \le \pi$.

1. Expand the terms on the left-hand side using the addition formulae for sine.

$\text{LHS} = \left(\sin\frac{\pi}{3}\cos x + \cos\frac{\pi}{3}\sin x\right) - \left(\sin\frac{\pi}{3}\cos x - \cos\frac{\pi}{3}\sin x\right)$

2. Simplify the expression by cancelling the common terms.

$\text{LHS} = \sin\frac{\pi}{3}\cos x + \cos\frac{\pi}{3}\sin x - \sin\frac{\pi}{3}\cos x + \cos\frac{\pi}{3}\sin x$
$= 2\cos\frac{\pi}{3}\sin x$

3. Substitute the exact value of $\cos\frac{\pi}{3}$ to complete the proof.

Since $\cos\frac{\pi}{3} = \frac{1}{2}$,
$\text{LHS} = 2\left(\frac{1}{2}\right)\sin x$
$= \sin x = \text{RHS}$

1. Apply the identity from part (a) to simplify the equation

From part (a), $\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)=\sin x$.
Replacing $x$ with $2x$, we get:
$\sin\left(\frac{\pi}{3}+2x\right)-\sin\left(\frac{\pi}{3}-2x\right) = \sin 2x$
Substitute this into the given equation:
$\sin 2x - 2 = 4\sin 2x$

2. Rearrange the equation to make $\sin 2x$ the subject

$\sin 2x - 4\sin 2x = 2$
$-3\sin 2x = 2$
$\sin 2x = -\frac{2}{3}$

3. Find the basic angle and determine the relevant quadrants

Basic angle $\alpha = \sin^{-1}\left(\frac{2}{3}\right) \approx 0.72973$ rad
Since $0 \le x \le \pi$, the range for $2x$ is $0 \le 2x \le 2\pi$.
Since $\sin 2x$ is negative, $2x$ lies in the 3rd and 4th quadrants.

4. Solve for $2x$ and then $x$

$2x = \pi + 0.72973$ or $2x = 2\pi - 0.72973$
$2x = 3.8713$ or $5.5535$
$x = 1.9356$ or $2.7767$
$x \approx 1.94$ or $2.78$ (to 3 sig. fig.)