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: Rewrite the equation.
\[ \sqrt{3}\sec x + 2 = 0 \quad\Rightarrow\quad \sec x = -\frac{2}{\sqrt{3}} \]
Step 2: Turn sec into cos.
Since $\sec x = \frac{1}{\cos x}$, take the reciprocal.
\[ \cos x = -\frac{\sqrt{3}}{2} \]
Step 3: Find the reference angle.
$\cos x = \frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{6}$. So the reference angle is $\frac{\pi}{6}$.

Step 4: Pick the right quadrants.
Cosine is negative in the second and third quadrants.

Step 5: Write the principal solutions.
Second quadrant: $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Third quadrant: $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

Step 6: Conclusion.
Both lie between $0$ and $2\pi$, so they are the principal solutions. \[ \boxed{\frac{5\pi}{6},\ \frac{7\pi}{6} \text{ (Option 2)}} \]