Solve the equation $2 \cot^2 2x = 1 - \frac{5}{\sin 2x}$ for $-180^\circ \le x \le 180^\circ$.

1. Express the equation entirely in terms of $\sin 2x$.

$2 \cot^2 2x = 1 - \frac{5}{\sin 2x}$
$2 \left( \frac{\cos^2 2x}{\sin^2 2x} \right) = 1 - \frac{5}{\sin 2x}$
$2 \left( \frac{1 - \sin^2 2x}{\sin^2 2x} \right) = 1 - \frac{5}{\sin 2x}$

2. Form a quadratic equation in $\sin 2x$.

Multiplying throughout by $\sin^2 2x$ (since $\sin 2x \neq 0$):
$2(1 - \sin^2 2x) = \sin^2 2x - 5\sin 2x$
$2 - 2\sin^2 2x = \sin^2 2x - 5\sin 2x$
$3\sin^2 2x - 5\sin 2x - 2 = 0$

3. Solve the quadratic equation for $\sin 2x$.

$(3\sin 2x + 1)(\sin 2x - 2) = 0$
$\sin 2x = -\frac{1}{3} \quad \text{or} \quad \sin 2x = 2$
Since $-1 \le \sin 2x \le 1$, $\sin 2x = 2$ has no real solution.

4. Find the basic angle and the range for $2x$.

Basic angle $\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.471^\circ$
Given $-180^\circ \le x \le 180^\circ$, the range for $2x$ is $-360^\circ \le 2x \le 360^\circ$.

5. Find all possible values for $2x$ in the required range.

Since $\sin 2x$ is negative, $2x$ lies in the 3rd and 4th quadrants.
For $0^\circ \le 2x \le 360^\circ$:
$2x = 180^\circ + 19.471^\circ = 199.471^\circ$
$2x = 360^\circ - 19.471^\circ = 340.529^\circ$
For $-360^\circ \le 2x < 0^\circ$:
$2x = -180^\circ + 19.471^\circ = -160.529^\circ$
$2x = -19.471^\circ$

6. Calculate the final values of $x$.

$x = \frac{199.471^\circ}{2}, \frac{340.529^\circ}{2}, \frac{-160.529^\circ}{2}, \frac{-19.471^\circ}{2}$
$x = 99.7^\circ, 170.3^\circ, -80.3^\circ, -9.7^\circ \text{ (to 1 d.p.)}$