Given that $\frac{\sin(A + B)}{\sin(A - B)} = \frac{1}{5}$, prove that $2\tan A + 3\tan B = 0$. Hence, solve the equation $\sin(x - 45^\circ) = 5\sin(x + 45^\circ)$ for $0^\circ < x < 360^\circ$.

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

$5\sin(A + B) = \sin(A - B)$

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

$5(\sin A \cos B + \cos A \sin B) = \sin A \cos B - \cos A \sin B$

3. Expand the brackets and collect like terms.

$5\sin A \cos B + 5\cos A \sin B = \sin A \cos B - \cos A \sin B$
$4\sin A \cos B + 6\cos A \sin B = 0$

4. Divide the equation by $\cos A \cos B$ to express in terms of tangent, and simplify.

$\frac{4\sin A \cos B}{\cos A \cos B} + \frac{6\cos A \sin B}{\cos A \cos B} = 0$
$4\tan A + 6\tan B = 0$
$2\tan A + 3\tan B = 0 \quad \text{(Shown)}$

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

$\sin(x - 45^\circ) = 5\sin(x + 45^\circ)$
$\frac{\sin(x + 45^\circ)}{\sin(x - 45^\circ)} = \frac{1}{5}$

2. Apply the result from part (i) to form an equation in terms of tan

Using the result from (i) with $A = x$ and $B = 45^\circ$:
$2\tan x + 3\tan 45^\circ = 0$

3. Evaluate $\tan 45^\circ$ and solve for $\tan x$

Since $\tan 45^\circ = 1$:
$2\tan x + 3(1) = 0$
$\tan x = -\frac{3}{2} = -1.5$

4. Find the basic angle and solve for $x$ in the given domain

Basic angle $\alpha = \tan^{-1}(1.5) \approx 56.3099^\circ$
Since $\tan x$ is negative, $x$ lies in the 2nd and 4th quadrants.
$x = 180^\circ - 56.3099^\circ = 123.6901^\circ \approx 123.7^\circ$
$x = 360^\circ - 56.3099^\circ = 303.6901^\circ \approx 303.7^\circ$