Question

The principal solutions of $\sqrt{3} \sec x + 2 = 0$ are

• A. $\frac{\pi}{6}, \frac{5\pi}{6}$
• B. $\frac{5\pi}{6}, \frac{7\pi}{6}$
• C. $\frac{\pi}{3}, \frac{2\pi}{3}$
• D. $\frac{2\pi}{3}, \frac{4\pi}{3}$

Hint:

Always convert $\sec, \csc,$ and $\cot$ into $\cos, \sin,$ and $\tan$ immediately. It makes finding the reference angle much easier since we have standard values memorized for the primary trigonometric ratios.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the principal solutions (angles between $0$ and $2\pi$) for the given trigonometric equation.

Step 2: Detailed Explanation:
The given equation is $\sqrt{3} \sec x + 2 = 0$.
Rearrange to isolate $\sec x$:
$$\sec x = -\frac{2}{\sqrt{3}}$$ Convert secant to cosine by taking the reciprocal:
$$\cos x = -\frac{\sqrt{3}}{2}$$ The cosine function is negative in the second and third quadrants.
The reference angle (where $\cos \alpha = \frac{\sqrt{3}}{2}$) is $\alpha = \frac{\pi}{6}$.
For the second quadrant solution:
$$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ For the third quadrant solution:
$$x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$

Step 3: Final Answer:
The principal solutions are $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$, matching option (B).