Find the general value of $\theta$ for the inverse trigonometric function $\text { Find the general value of } \theta, \text { when } 9 \sec ^{4} \theta=16$
My work-
Given $\sec ^{4} \theta=\frac{16}{9}$
or, $\sec ^{4} \theta = \frac{(4)^{2}}{(3)^{2}}$
$\implies \sec ^{2} \theta=\frac{4}{3} \Longrightarrow \theta=\sec ^{-1}\left(\sqrt{\frac{4}{3}}\right)$
hence, $\theta=2 n \pi \pm \frac{\pi}{6}$.
I am following Hobson's Trigonometry and it doesn't have the answer or the solution provided, neither any examples regarding inverse trigonometric functions. I got the feedback that my approach is incorrect but I cannot see how to fix it so any hints on where I am going mistaken ?
P.S. Hobson mentions that $x = \sec^{-1}(y)$ is a multi-valued function.

$$
\begin{aligned}
9 \sec ^{4} \theta &=16 \\
\sec ^{4} \theta &=\frac{16}{9} \\
\sec \theta &=\pm \frac{2}{\sqrt{3}} \\
\cos \theta &=\pm \frac{\sqrt{3}}{2} \\
\theta &=n \pi \pm \frac{\pi}{6}=\frac{(6 n \pm 1) \pi}{6},
\end{aligned}
$$
where $n \in \mathbb{Z}.$
