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 $\pi$?
 A: 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*}
A: $$\sec^2x=\dfrac43 \iff\cos^2x=\dfrac1{\sec^2x}=?$$
$$\iff\sin^2x=\cdots=\dfrac14=\sin^2\dfrac\pi6$$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin\left(x-\dfrac\pi6\right)\cdot\sin\left(x+\dfrac\pi6\right)=0$$
If $\sin\left(x-\dfrac\pi6\right)=0, x-\dfrac\pi6=m\pi$
If $\sin\left(x+\dfrac\pi6\right)=0, x=-\dfrac\pi6+m\pi=(m-1)\pi+\left(\pi-\dfrac\pi6\right)$
where $m$ is ay integer
A: Sometimes a plot can really help:

A: WolframAlpha offers $2$ solutions here
where
$$x = \pi n - \frac{5 \pi}{6}, n \in\mathbb{Z}$$
$$x = \pi n - \frac{\pi}{6}, n \in\mathbb{Z}$$
Your addition of $2n\pi$ is always a multiple of $360^\circ$ so the results would be the same but $\quad n\pi\quad$ would take you, for example from quadrant $(1)$ to quadrant $(3)$ where the $\sin$ and $\cos$ are both negative. In any case the $\sec^2$ would still be  positive so a rotation of only $180^ \circ$ will not affect the solution.
