How do we solve the equation $3\sec^{2}(x) - 4 = 0$?

Question

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 $\pi$?

Answer 1

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*}

Answer 2

Sometimes a plot can really help:

Answer 3

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.

Answer 4

$$\sec^2x=\dfrac43 \iff\cos^2x=\dfrac1{\sec^2x}=?$$

$$\iff\sin^2x=\cdots=\dfrac14=\sin^2\dfrac\pi6$$

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