Use inverse functions where needed to find all solutions of the equation in the interval [0,2π). $\sec ^ { 2 } x - 6 \sec x = 0$
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 [0,2π).​ $\sec ^ { 2 } x - 6 \sec x = 0$ ​
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

The equation $\sec^2x - 6\sec x = 0$ can be solved by factoring: $\sec x(\sec x - 6) = 0$, leading to $\sec x = 0$ (no solution, as secant never equals zero) and $\sec x = 6$. The secant function is the reciprocal of the cosine function, so $\cos x = \frac{1}{6}$. The solutions to $\cos x = \frac{1}{6}$ in the interval [0, 2π) are given by $x = \arccos\left(\frac{1}{6}\right)$ and $x = 2\pi - \arccos\left(\frac{1}{6}\right)$, as cosine is positive in the first and fourth quadrants.

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