Answer: \(x=0,\ \pm\dfrac{\pi}{4},\ \pm\dfrac{3\pi}{4}\).

We start with the main method. Factorise the trigonometric equation first, then solve each factor over the required interval.

For part (a), use

\(\csc x=\dfrac{1}{\sin x}\quad\text{and}\quad \cot x=\dfrac{\cos x}{\sin x}\).

Then

\(\displaystyle \sin^3x\left(\dfrac{\csc x}{\cot x}\right)\)

\(\displaystyle =\sin^3x\left(\dfrac{1/\sin x}{\cos x/\sin x}\right)\).

This simplifies to

\(\displaystyle \sin^3x\cdot\dfrac{1}{\cos x}\)

\(\displaystyle =\sin^2x\cdot\dfrac{\sin x}{\cos x}\).

Hence

\(\displaystyle \sin^3x\left(\dfrac{\csc x}{\cot x}\right)=\sin^2x\tan x\).

For part (b), factorise without dividing by \(\tan x\):

\(\displaystyle \cos^2x\tan x-\dfrac{1}{2}\tan x=0\)

\(\displaystyle \tan x\left(\cos^2x-\dfrac{1}{2}\right)=0\).

So either

\(\tan x=0\)

or

\(\cos^2x=\dfrac{1}{2}\).

From \(\tan x=0\), in \(-\pi<x<\pi\),

\(x=0\).

From \(\cos^2x=\dfrac{1}{2}\),

\(\cos x=\pm\dfrac{1}{\sqrt2}\).

So in the interval \(-\pi<x<\pi\),

\(x=\pm\dfrac{\pi}{4},\quad \pm\dfrac{3\pi}{4}\).

Therefore the complete set of solutions is

\(x=0,\ \pm\dfrac{\pi}{4},\ \pm\dfrac{3\pi}{4}\).

This completes the solution and gives the required result.