Question

The general solution of \[ 2\cos\theta-\sqrt3=0 \] is

• A. \(\theta=n\pi+(-1)^n\frac{\pi}{6},\ n\in\mathbb{Z}\)
• B. \(\theta=2n\pi\pm\frac{\pi}{6},\ n\in\mathbb{Z}\)
• C. \(\theta=2n\pi\pm\frac{\pi}{3},\ n\in\mathbb{Z}\)
• D. \(\theta=n\pi+(-1)^n\frac{\pi}{3},\ n\in\mathbb{Z}\)

Hint:

Remember: \[ \cos\theta=\cos\alpha \Rightarrow \theta=2n\pi\pm\alpha. \] \[ \sin\theta=\sin\alpha \Rightarrow \theta=n\pi+(-1)^n\alpha. \] Students often interchange these two formulas.

Answer 1

The Correct Option is B

Step 1: Simplify the equation. Given, \[ 2\cos\theta-\sqrt3=0. \] Adding \(\sqrt3\) to both sides gives \[ 2\cos\theta=\sqrt3. \] Dividing by \(2\), \[ \cos\theta=\frac{\sqrt3}{2}. \]

Step 2: Find the principal angle. We know that \[ \cos\frac{\pi}{6} = \frac{\sqrt3}{2}. \] Therefore, \[ \alpha=\frac{\pi}{6}. \]

Step 3: Write the general solution. Using \[ \theta=2n\pi\pm\alpha, \] we obtain \[ \theta = 2n\pi \pm \frac{\pi}{6}, \quad n\in\mathbb Z. \] Conclusion: Hence, \[ {\theta=2n\pi\pm\frac{\pi}{6},\quad n\in\mathbb Z}. \]