Answer: \(f(x)=3\tan^2x+1\); \(x=\pm\dfrac{\pi}{4}\); gradients \(12\) and \(-12\); \(\theta=11.5^\circ,168.5^\circ,197.5^\circ,342.5^\circ\).

We start with the main method. Use the product, chain and standard trigonometric derivative rules carefully.

First rewrite the function:

\(\mathrm f(x)=\frac{1+2\sin^2x}{\cos^2x}=\frac1{\cos^2x}+2\frac{\sin^2x}{\cos^2x}.\)

So

\(\mathrm f(x)=\sec^2x+2\tan^2x.\)

Using \(\sec^2x=1+\tan^2x\),

\(\mathrm f(x)=3\tan^2x+1.\)

Now solve \(\mathrm f(x)=4\):

\(3\tan^2x+1=4.\)

Thus \(\tan^2x=1\), so \(\tan x=\pm1\).

Since \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\), the solutions are

\(x=-\frac{\pi}{4}\) and \(x=\frac{\pi}{4}\).

Differentiate \(3\tan^2x+1\):

\(\mathrm f'(x)=6\tan x\sec^2x.\)

At \(x=\frac{\pi}{4}\), \(\tan x=1\) and \(\sec^2x=2\), so

\(\mathrm f'\left(\frac{\pi}{4}\right)=12.\)

At \(x=-\frac{\pi}{4}\), \(\tan x=-1\) and \(\sec^2x=2\), so

\(\mathrm f'\left(-\frac{\pi}{4}\right)=-12.\)

For part (b), use \(\cos^2\theta=1-\sin^2\theta\):

\(50(1-\sin^2\theta)=5\sin\theta+47.\)

This simplifies to

\(50\sin^2\theta+5\sin\theta-3=0.\)

Factorising,

\((10\sin\theta+3)(5\sin\theta-1)=0.\)

So \(\sin\theta=-0.3\) or \(\sin\theta=0.2\).

In the interval \(0^\circ\leqslant\theta\leqslant360^\circ\), this gives

\(\theta=11.5^\circ,\ 168.5^\circ,\ 197.5^\circ,\ 342.5^\circ\) to 1 decimal place.

This completes the solution and gives the required result.