Solve the following equation. $9 \sec ^ { 2 } x - 12 = 0$
A)$\frac { 7 \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$
B)$\frac { \pi } { 6 } + n \pi , \frac { 5 \pi } { 6 } + n \pi$
C) $\frac { \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$
D) $\frac { 5 \pi } { 2 } + 2 \pi , \frac { 5 \pi } { 2 } + 2 \pi$
E) $\frac { 3 \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$

Question

Solve the following equation. $9 \sec ^ { 2 } x - 12 = 0$ 
A)$\frac { 7 \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$
B)$\frac { \pi } { 6 } + n \pi , \frac { 5 \pi } { 6 } + n \pi$
C) $\frac { \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$
D) $\frac { 5 \pi } { 2 } + 2 \pi , \frac { 5 \pi } { 2 } + 2 \pi$
E) $\frac { 3 \pi } { 2 } + 2 \pi , \frac { 3 \pi } { 2 } + 2 \pi$

Quizplus · Accepted Answer

First, solve the equation $9\sec^2x - 12 = 0$ for $\sec^2x$, which gives $\sec^2x = \frac{12}{9} = \frac{4}{3}$. Taking the square root gives $\sec x = \pm\sqrt{\frac{4}{3}} = \pm\frac{2}{\sqrt{3}}$, which simplifies to $\sec x = \pm\frac{2\sqrt{3}}{3}$. Recognizing that $\sec x = \frac{1}{\cos x}$, we have $\cos x = \pm\frac{\sqrt{3}}{2}$. The solutions for $\cos x = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{6} + 2\pi n$ and $x = -\frac{\pi}{6} + 2\pi n$, but since we are looking for all solutions, we also consider the negative value of cosine, which leads to $x = \frac{5\pi}{6} + 2\pi n$ and $x = -\frac{5\pi}{6} + 2\pi n$. However, the general form for all solutions of cosine equations is $x = \frac{\pi}{6} + n\pi$ and $x = \frac{5\pi}{6} + n\pi$, where $n$ is any integer, to account for the periodic nature of the cosine function in both positive and negative directions on the unit circle.

Solve the Following Equation 9sec⁡2x−12=09 \sec ^ { 2 } x - 12 = 09sec2x−12=0

Solve the Following Equation $9 \sec ^ { 2 } x - 12 = 0$