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: Understanding the Question:
The question asks for the principal solutions of the trigonometric equation $\cot x = \sqrt{3}$. Principal solutions are the angle values of $x$ that satisfy the equation within the restricted domain interval $[0, 2\pi)$.

Step 2: Key Formula or Approach:
We can rewrite the cotangent function in terms of its reciprocal tangent relation:
$$\cot x = \sqrt{3} \implies \tan x = \frac{1}{\cot x} = \frac{1}{\sqrt{3}}$$ Since $\tan x$ is positive, solutions must lie in the first quadrant ($\text{Q}_1$) and the third quadrant ($\text{Q}_3$), where tangent is positive.

Step 3: Detailed Explanation:
Let's find the reference angle in the first quadrant:
We know from standard trigonometric tables that $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. Thus, the first principal solution is:
$$x = \frac{\pi}{6}$$ Now, use the trigonometric identity for the third quadrant, $\tan(\pi + \theta) = \tan \theta$, to find the second solution:
$$x = \pi + \frac{\pi}{6} = \frac{6\pi + \pi}{6} = \frac{7\pi}{6}$$ Let's verify that both values fall inside our boundary criteria:
$0 \le \frac{\pi}{6} < 2\pi$ (Valid)
$0 \le \frac{7\pi}{6} < 2\pi$ (Valid)
Therefore, the two principal angles are $\frac{\pi}{6}$ and $\frac{7\pi}{6}$, matching option (C).

Step 4: Final Answer:
The principal solutions are $\frac{\pi}{6}$ and $\frac{7\pi}{6}$, which corresponds to option (C).