Answer: \(x=15^\circ,38.9^\circ,75^\circ,98.9^\circ\).
Use the identity
\(\sec^2\theta=1+\tan^2\theta\).
With \(\theta=3x\), the equation becomes
\(1+\tan^2 3x+\tan3x-3=0\).
So
\(\tan^2 3x+ an3x-2=0\).
Factorising,
\((\tan3x+2)(\tan3x-1)=0\).
Hence
\(\tan3x=1\quad\text{or}\quad\tan3x=-2\).
Since
\(0^\circ\leqslant x\leqslant120^\circ\),
we have
\(0^\circ\leqslant3x\leqslant360^\circ\).
For \(\tan3x=1\), the solutions in this interval are
\(3x=45^\circ,225^\circ\).
Therefore
\(x=15^\circ,75^\circ\).
For \(\tan3x=-2\), the reference angle is
\(\tan^{-1}2=63.4349\ldots^\circ\).
Since tangent is negative in quadrants II and IV,
\(3x=180^\circ-63.4349\ldots^\circ=116.565\ldots^\circ\),
or
\(3x=360^\circ-63.4349\ldots^\circ=296.565\ldots^\circ\).
Thus
\(x=38.855\ldots^\circ\quad\text{or}\quad x=98.855\ldots^\circ\).
To 3 significant figures,
\(x=15^\circ,38.9^\circ,75^\circ,98.9^\circ\).
