Given that $\dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{2}{3}$, prove that $\tan A = 5\tan B$. Hence, or otherwise, solve the equation $3\sin(A-45^\circ)=2\sin(A+45^\circ)$ for $0^\circ \le A \le 360^\circ$.

1. Cross-multiply the given equation to remove fractions.

$3\sin(A-B) = 2\sin(A+B)$

2. Expand both sides using the addition formulae for sine.

$3(\sin A \cos B - \cos A \sin B) = 2(\sin A \cos B + \cos A \sin B)$

3. Expand the brackets and collect like terms.

$3\sin A \cos B - 3\cos A \sin B = 2\sin A \cos B + 2\cos A \sin B$
$3\sin A \cos B - 2\sin A \cos B = 2\cos A \sin B + 3\cos A \sin B$
$\sin A \cos B = 5\cos A \sin B$

4. Divide both sides by $\cos A \cos B$ to obtain the required tangent terms.

$\frac{\sin A \cos B}{\cos A \cos B} = \frac{5\cos A \sin B}{\cos A \cos B}$
$\tan A = 5\tan B$

1. Rearrange the given equation to match the form in part (i)

$3\sin(A-45^\circ)=2\sin(A+45^\circ)$
$\frac{\sin(A-45^\circ)}{\sin(A+45^\circ)} = \frac{2}{3}$

2. Apply the result from part (i) to simplify the equation

Using the result from (i) with $B = 45^\circ$:
$\tan A = 5\tan 45^\circ$
Since $\tan 45^\circ = 1$,
$\tan A = 5(1) = 5$

3. Solve for A in the given interval

Basic angle $\alpha = \tan^{-1}(5) \approx 78.690^\circ$
Since $\tan A$ is positive, $A$ lies in the 1st and 3rd quadrants.
$A = 78.690^\circ \text{ or } 180^\circ + 78.690^\circ$
$A = 78.7^\circ \text{ or } 258.7^\circ \text{ (to 1 d.p.)}$