Solve the following equation. $9 \cot ^ { 2 } x - 3 = 0$
A) $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
B) $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
C) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
D) $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E) $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$

Question

Solve the following equation.   $9 \cot ^ { 2 } x - 3 = 0$  
A)  $\frac { \pi } { 2 } + n \pi , \frac { 2 \pi } { 5 } + n \pi$
B)  $\frac { \pi } { 3 } + n \pi , \frac { \pi } { 3 } + n \pi$
C) $\frac { 4 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
D)  $\frac { 2 \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$
E) $\frac { \pi } { 3 } + n \pi , \frac { 2 \pi } { 3 } + n \pi$

Quizplus · Accepted Answer

First, solve the equation $9 \cot ^ { 2 } x - 3 = 0$ for $\cot^2 x$. Adding 3 to both sides and then dividing by 9 gives $\cot^2 x = \frac{1}{3}$. Taking the square root of both sides, we get $\cot x = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$. The cotangent function equals $\frac{1}{\sqrt{3}}$ at $x = \frac{\pi}{6}$ and $-\frac{1}{\sqrt{3}}$ at $x = \frac{5\pi}{6}$, but since cotangent has a period of $\pi$, the general solutions are $x = \frac{\pi}{6} + n\pi$ and $x = \frac{5\pi}{6} + n\pi$, which correspond to $\frac{\pi}{3} + n\pi$ and $\frac{2\pi}{3} + n\pi$ respectively, due to the symmetry of the cotangent function in its period.

Solve the Following Equation 9cot⁡2x−3=09 \cot ^ { 2 } x - 3 = 09cot2x−3=0

Solve the Following Equation $9 \cot ^ { 2 } x - 3 = 0$