Answer: \(x=0^\circ,45^\circ,135^\circ,180^\circ\); \(\theta=-\dfrac{5\pi}{9},-\dfrac{\pi}{9},\dfrac{\pi}{9},\dfrac{5\pi}{9}\).

For part (a), since \(\tan x=\dfrac{\sin x}{\cos x}\), the equation is

\(\sin x\cos x=\frac{\sin x}{2\cos x}.\)

Multiplying by \(2\cos x\),

\(2\sin x\cos^2x=\sin x.\)

Hence

\(\sin x(2\cos^2x-1)=0.\)

So \(\sin x=0\), giving \(x=0^\circ,180^\circ\), or \(\cos^2x=\frac12\), giving \(x=45^\circ,135^\circ\).

Therefore

\(x=0^\circ,\ 45^\circ,\ 135^\circ,\ 180^\circ.\)

For the identity,

\(\frac{\sin\theta}{\operatorname{cot}\theta} =\frac{\sin\theta}{\cos\theta/\sin\theta} =\frac{\sin^2\theta}{\cos\theta}.\)

Thus

\(\operatorname{sec}\theta-\frac{\sin\theta}{\operatorname{cot}\theta} =\frac1{\cos\theta}-\frac{\sin^2\theta}{\cos\theta} =\frac{1-\sin^2\theta}{\cos\theta} =\frac{\cos^2\theta}{\cos\theta} =\cos\theta.\)

Using the identity with angle \(3\theta\), the equation becomes

\(\cos3\theta=\frac12.\)

Let \(\phi=3\theta\). Since \(-\frac{2\pi}{3}\leq\theta\leq\frac{2\pi}{3}\),

\(-2\pi\leq\phi\leq2\pi.\)

In this interval,

\(\phi=-\frac{5\pi}{3},\ -\frac{\pi}{3},\ \frac{\pi}{3},\ \frac{5\pi}{3}.\)

Dividing by 3,

\(\boxed{\theta=-\frac{5\pi}{9},\ -\frac{\pi}{9},\ \frac{\pi}{9},\ \frac{5\pi}{9}}.\)