Show that $\dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}=\dfrac{2}{\sin x}$. Hence solve the equation $\dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}=1+3\sin x$ for $0^\circ \le x \le 360^\circ$.

1. Express the left-hand side (LHS) in terms of sine and cosine.

$\text{LHS} = \frac{\frac{\sin x}{\cos x}}{1+\frac{1}{\cos x}} + \frac{1+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}}$

2. Simplify the complex fractions by multiplying numerator and denominator by $\cos x$.

$\text{LHS} = \frac{\sin x}{\cos x + 1} + \frac{\cos x + 1}{\sin x}$

3. Combine the two fractions into a single fraction with a common denominator.

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

4. Expand the numerator and apply the Pythagorean identity $\sin^2 x + \cos^2 x = 1$.

5. Factorise the numerator and cancel the common factor to obtain the right-hand side (RHS).

1. Substitute the result from part (i) into the equation.

Using the result from (i), the equation becomes:
$\frac{2}{\sin x} = 1 + 3\sin x$

2. Rearrange into a quadratic equation in terms of $\sin x$.

$2 = \sin x + 3\sin^2 x$
$3\sin^2 x + \sin x - 2 = 0$

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

$(3\sin x - 2)(\sin x + 1) = 0$
$\sin x = \frac{2}{3} \quad \text{or} \quad \sin x = -1$

4. Find the values of $x$ in the given range, rejecting any invalid solutions.

For $\sin x = \frac{2}{3}$:
Basic angle $\alpha = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.810^\circ$
Since $\sin x$ is positive, $x$ is in the 1st and 2nd quadrants:
$x = 41.810^\circ \quad \text{or} \quad x = 180^\circ - 41.810^\circ = 138.190^\circ$

For $\sin x = -1$:
$x = 270^\circ$
However, at $x = 270^\circ$, $\tan x$ and $\sec x$ are undefined, so $x = 270^\circ$ must be rejected.

$x \approx 41.8^\circ, 138.2^\circ \text{ (to 1 d.p.)}$