Answer: \(x=-110^\circ,-70^\circ,10^\circ,50^\circ,130^\circ,170^\circ\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

Use \(1+\tan^2\theta=\sec^2\theta=\frac{1}{\cos^2\theta}\) and \(\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}\).

Then

\(\frac{\sin\theta\tan^2\theta}{1+\tan^2\theta}=\frac{\sin\theta\cdot\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}=\sin^3\theta.\)

Using this result with \(\theta=3x\), the equation becomes

\(\sin^3 3x=\frac18.\)

So

\(\sin3x=\frac12.\)

Since \(-180^\circ\leqslant x\leqslant180^\circ\),

\(-540^\circ\leqslant3x\leqslant540^\circ.\)

The solutions of \(\sin3x=\frac12\) in this interval are

\(3x=-330^\circ,-210^\circ,30^\circ,150^\circ,390^\circ,510^\circ.\)

Therefore

\(x=-110^\circ,-70^\circ,10^\circ,50^\circ,130^\circ,170^\circ.\)

This gives the required answer: \(x=-110^\circ,-70^\circ,10^\circ,50^\circ,130^\circ,170^\circ\).