Use inverse functions where needed to find all solutions of the equation in the interval [0,2π). $\cot ^ { 2 } x - 25 = 0$
A) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi$
B) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi$
C) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$
D) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$
E) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$

Question

Use inverse functions where needed to find all solutions of the equation in the interval [0,2π).​ $\cot ^ { 2 } x - 25 = 0$ ​
A) $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi$
B)​ $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( \frac { 1 } { 5 } \right) + \pi$
C)​ $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$
D)​ $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$
E)​ $\arctan \left( \frac { 1 } { 5 } \right) , \arctan \left( \frac { 1 } { 5 } \right) + 2 \pi , \arctan \left( - \frac { 1 } { 5 } \right) + \pi , \arctan \left( - \frac { 1 } { 5 } \right) + 2 \pi$

Quizplus · Accepted Answer

First, solve the equation $\cot^2 x = 25$, which gives $\cot x = \pm 5$. The solutions for $\cot x = 5$ and $\cot x = -5$ can be found using the inverse cotangent function, but since cotangent is the reciprocal of tangent, we use $\arctan(\frac{1}{5})$ and $\arctan(-\frac{1}{5})$ to find the principal values. For $\cot x = 5$, the solution is $\arctan(\frac{1}{5})$. Since cotangent has a period of $\pi$, the next solution within the interval [0, 2π) is $\arctan(\frac{1}{5}) + \pi$. For $\cot x = -5$, the solutions are $\arctan(-\frac{1}{5}) + \pi$ and $\arctan(-\frac{1}{5}) + 2\pi$, but since the interval is [0, 2π), we only consider up to $2\pi$, making the last term redundant. Therefore, the correct solutions are those that account for the periodicity of cotangent and the given interval, which are found in option C.

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