<s>[INST] How do we solve the equation $3\sec^{2}(x) - 4 = 0$? I am confused about the answer to this question:
$$3\sec^{2}(x) - 4 = 0$$
I got the answer
$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$
However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$
Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?
 [/INST] The proposed equation is equivalent to
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow \cos(x) = \pm\frac{\sqrt{3}}{2}\\\\
& \Longleftrightarrow \cos(x) = \pm\cos\left(\frac{\pi}{6}\right)
\end{align*}
Based on it, we can conclude that
\begin{align*}
\cos(x) = \cos\left(\frac{\pi}{6}\right) \Longleftrightarrow x = \pm\frac{\pi}{6} + 2m\pi
\end{align*}
as well as
\begin{align*}
\cos(x) = -\cos\left(\frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \Longleftrightarrow x = \pm\frac{5\pi}{6} + 2n\pi
\end{align*}
Gathering all solutions, we obtain the following solution set:
\begin{align*}
S = \left\{x\in\mathbb{R} \mid \left(x = \frac{\pi}{6} + m\pi\right)\vee\left(x = \frac{5\pi}{6} + n\pi\right)\right\}
\end{align*}
In order to understand it properly, notice that
\begin{align*}
\begin{cases}
-\dfrac{5\pi}{6} = \dfrac{\pi}{6} - \pi\\\\
-\dfrac{\pi}{6} = \dfrac{5\pi}{6} - \pi
\end{cases}
\end{align*}
Based on it, we can express the solution set as proposed in the book.
EDIT
Considering the comment of @HansEngler, we can also solve the proposed equation as follows:
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow 2\cos(2x) - 1 = 0\\\\\
& \Longleftrightarrow \cos(2x) = \cos\left(\frac{\pi}{3}\right)\\\\
& \Longleftrightarrow 2x = \pm\frac{\pi}{3} + 2k\pi\\\\
& \Longleftrightarrow x = \pm\frac{\pi}{6} + k\pi
\end{align*}
</s>