Question

The common principal solution of the equations $\sin \theta = -1/2$ and $\tan \theta = 1/\sqrt{3}$ is \dots}

• A. $\pi/6$
• B. $5\pi/6$
• C. $7\pi/6$
• D. $11\pi/6$

Hint:

The ASTC (All Students Take Calculus) mnemonic is foolproof for these problems. Quadrant 1 (All +), Quadrant 2 (Sin +), Quadrant 3 (Tan +), Quadrant 4 (Cos +). By combining the sign constraints, the quadrant isolates itself instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must find a single angle $\theta$ in the principal range $[0, 2\pi)$ that simultaneously satisfies two distinct trigonometric equations.

Step 2: Key Formula or Approach:
Use the signs (positive or negative) of the given trigonometric values to determine the correct quadrant in the Cartesian plane (ASTC rule). Then find the reference angle and map it to that quadrant.

Step 3: Detailed Explanation:
1. Analyze the given values:
- $\sin \theta = -1/2$. The sine function is negative in Quadrant III and Quadrant IV.
- $\tan \theta = 1/\sqrt{3}$. The tangent function is positive in Quadrant I and Quadrant III.
2. Determine the common quadrant:
The only quadrant where both conditions can be true simultaneously is Quadrant III.
3. Find the base (reference) angle:
Look at the absolute numerical values: $\sin(\alpha) = 1/2$ and $\tan(\alpha) = 1/\sqrt{3}$.
The basic acute angle that satisfies these is $\alpha = \pi/6$ (or $30^\circ$).
4. Map the reference angle to Quadrant III:
In Quadrant III, the angle is expressed as $\pi + \alpha$.
$$\theta = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6}$$

Step 4: Final Answer:
The common principal solution is $7\pi/6$, matching option (c).