Question

Which of the following is not a solution of the equation $3\tan^2\theta - \sin\theta = 0$?

• A. $n\pi$
• B. $n\pi + \frac{\pi}{2}$
• C. $n\pi + (-1)^n\frac{\pi}{6}$
• D. $0$
• E. $\pi$

Hint:

Always check domain restrictions when $\tan$ is involved.

Answer 1

The Correct Option is B

Step 1: Rewrite equation.
\[ 3\tan^2\theta - \sin\theta = 0 \] \[ 3\frac{\sin^2\theta}{\cos^2\theta} - \sin\theta = 0 \] Multiply by $\cos^2\theta$: \[ 3\sin^2\theta - \sin\theta \cos^2\theta = 0 \] \[ \sin\theta(3\sin\theta - \cos^2\theta)=0 \]

Step 2: Solve cases. Case 1: $\sin\theta=0$
\[ \theta = n\pi \] Case 2: \[ 3\sin\theta = \cos^2\theta = 1 - \sin^2\theta \] \[ \sin^2\theta + 3\sin\theta -1=0 \] Solve: \[ \sin\theta = \frac{-3 \pm \sqrt{9+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \] Valid solution: \[ \sin\theta = \frac{-3 + \sqrt{13}}{2} \] Corresponds to angle form: \[ n\pi + (-1)^n \alpha \]

Step 3: Check options.
$\tan\theta$ undefined at: \[ \theta = n\pi + \frac{\pi}{2} \] Thus not valid solution.
Final Answer: \[ \boxed{n\pi + \frac{\pi}{2}} \]