Question

The principal solutions of $\cot x = \sqrt{3}$ are

• A. $\frac{\pi}{6}, \frac{5\pi}{6}$
• B. $\frac{\pi}{4}, \frac{5\pi}{4}$
• C. $\frac{\pi}{6}, \frac{7\pi}{6}$
• D. $\frac{\pi}{3}, \frac{7\pi}{3}$

Hint:

Remember that the tangent and cotangent functions have a periodic cycle of exactly $\pi$. Once you find your primary first-quadrant angle ($\frac{\pi}{6}$), you can find the secondary principal solution simply by adding $\pi$ to it: $\frac{\pi}{6} + \pi = \frac{7\pi}{6}$!

Answer 1

The Correct Option is C

Step 1: Read what is being asked.
Principal solutions are all the angles $x$ in the interval $[0,2\pi)$ that satisfy $\cot x=\sqrt{3}$.
Step 2: Convert cotangent to tangent.
Since $\cot x=\dfrac{1}{\tan x}$, the equation $\cot x=\sqrt{3}$ becomes $\tan x=\dfrac{1}{\sqrt{3}}$.
Step 3: Decide the quadrants.
Here $\tan x$ is positive, and tangent is positive only in the first and third quadrants. So we expect one answer in each of those quadrants.
Step 4: Find the base angle.
From the standard table $\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$, so the reference angle is $\dfrac{\pi}{6}$. This is our first-quadrant solution, $x=\dfrac{\pi}{6}$.
Step 5: Get the third-quadrant solution.
In the third quadrant the angle is $\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6}$, and indeed $\tan\!\left(\dfrac{7\pi}{6}\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$.
Step 6: Check both lie in range.
Both $\dfrac{\pi}{6}$ and $\dfrac{7\pi}{6}$ sit inside $[0,2\pi)$, so these are exactly the principal solutions, which is option (3).
\[ \boxed{x=\dfrac{\pi}{6},\ \dfrac{7\pi}{6}} \]