Answer: \(\theta=-84.2^\circ,\ -5.8^\circ,\ 15^\circ,\ 75^\circ\).

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

Use \(\cot^2 2\theta=\operatorname{cosec}^2 2\theta-1\). The equation becomes

\(\operatorname{cosec}^2 2\theta+3\operatorname{cosec}2\theta-10=0.\)

Factorising,

\((\operatorname{cosec}2\theta-2)(\operatorname{cosec}2\theta+5)=0.\)

So \(\operatorname{cosec}2\theta=2\) or \(\operatorname{cosec}2\theta=-5\).

Hence \(\sin2\theta=\frac12\) or \(\sin2\theta=-\frac15\).

Since \(-90^\circ\leqslant\theta\leqslant90^\circ\), we have \(-180^\circ\leqslant2\theta\leqslant180^\circ\).

For \(\sin2\theta=\frac12\), \(2\theta=30^\circ,150^\circ\), so \(\theta=15^\circ,75^\circ\).

For \(\sin2\theta=-\frac15\), \(2\theta=-11.536\ldots^\circ,-168.464\ldots^\circ\), so \(\theta=-5.8^\circ,-84.2^\circ\).

This gives the required answer: \(\theta=-84.2^\circ,\ -5.8^\circ,\ 15^\circ,\ 75^\circ\).