Prove that $\sin^{3} x\sec^{2} x + \sin x = \tan x\sec x$. Solve the equation $\sin^{3} x\sec^{2} x + \sin x = 5$ for $0 \le x \le 360^\circ$.

1. Express the left-hand side (LHS) in terms of sine and cosine and combine into a single fraction.

$\text{LHS} = \sin^{3} x\left(\frac{1}{\cos^{2} x}\right) + \sin x$
$= \frac{\sin^{3} x + \sin x \cos^{2} x}{\cos^{2} x}$

2. Factorise the numerator and apply the Pythagorean identity.

$= \frac{\sin x (\sin^{2} x + \cos^{2} x)}{\cos^{2} x}$
$= \frac{\sin x (1)}{\cos^{2} x}$

3. Rewrite the expression to match the right-hand side (RHS).

$= \left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right)$
$= \tan x \sec x$
$= \text{RHS (shown)}$

1. Use the result from part (a) to simplify the equation

Using the identity from (a), the equation becomes:
$\tan x\sec x = 5$

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

$\left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right) = 5$
$\frac{\sin x}{\cos^2 x} = 5$
$\sin x = 5\cos^2 x$
$\sin x = 5(1 - \sin^2 x)$
$5\sin^2 x + \sin x - 5 = 0$

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

Using the quadratic formula:
$\sin x = \frac{-1 \pm \sqrt{1^2 - 4(5)(-5)}}{2(5)}$
$\sin x = \frac{-1 \pm \sqrt{101}}{10}$
$\sin x \approx 0.90499 \quad \text{or} \quad \sin x \approx -1.10499$
Since $-1 \le \sin x \le 1$, we reject $\sin x \approx -1.10499$.

4. Find the values of $x$ in the given range

Basic angle $\alpha = \sin^{-1}(0.90499) \approx 64.82^\circ$
Since $\sin x > 0$, $x$ lies in the 1st and 2nd quadrants.
$x = 64.82^\circ \quad \text{or} \quad x = 180^\circ - 64.82^\circ = 115.18^\circ$
$x = 64.8^\circ, 115.2^\circ \text{ (to 1 d.p.)}$