Use inverse functions where needed to find all solutions of the equation in the interval [0, 2$\pi $). $\sec ^ { 2 } x - 6 \sec x = 0$
A) $\arccos \left( \frac { 1 } { 6 } \right) , 2 \pi + \arccos \left( \frac { 1 } { 6 } \right)$
B) $\arccos \left( \frac { 1 } { 6 } \right) , \arccos \left( \frac { 1 } { 6 } \right) - 2 \pi$
C) $\arccos \left( \frac { 1 } { 6 } \right) + \pi , \arccos \left( \frac { 1 } { 6 } \right) - \pi$
D) $\arccos \left( \frac { 1 } { 6 } \right) , 2 \pi - \arccos \left( \frac { 1 } { 6 } \right)$
E) $\arccos \left( \frac { 1 } { 6 } \right) + 2 \pi , \arccos \left( \frac { 1 } { 6 } \right) - 2 \pi$

Question

Use inverse functions where needed to find all solutions of the equation in the interval [0, 2$\pi $).   $\sec ^ { 2 } x - 6 \sec x = 0$  
A) $\arccos \left( \frac { 1 } { 6 } \right) , 2 \pi + \arccos \left( \frac { 1 } { 6 } \right)$
B)  $\arccos \left( \frac { 1 } { 6 } \right) , \arccos \left( \frac { 1 } { 6 } \right) - 2 \pi$
C)  $\arccos \left( \frac { 1 } { 6 } \right) + \pi , \arccos \left( \frac { 1 } { 6 } \right) - \pi$
D)  $\arccos \left( \frac { 1 } { 6 } \right) , 2 \pi - \arccos \left( \frac { 1 } { 6 } \right)$
E)  $\arccos \left( \frac { 1 } { 6 } \right) + 2 \pi , \arccos \left( \frac { 1 } { 6 } \right) - 2 \pi$

Quizplus · Accepted Answer

First, factor the given equation: $\sec^2x - 6\sec x = 0$ can be factored as $\sec x(\sec x - 6) = 0$, giving $\sec x = 0$ and $\sec x = 6$. Since $\sec x = \frac{1}{\cos x}$, the equation $\sec x = 6$ becomes $\cos x = \frac{1}{6}$. The solutions to $\cos x = \frac{1}{6}$ in the interval $[0, 2\pi)$ are $\arccos\left(\frac{1}{6}\right)$ and $2\pi - \arccos\left(\frac{1}{6}\right)$, as cosine is positive in the first and fourth quadrants.

Use Inverse Functions Where Needed to Find All Solutions of the Equation