Question

If $6{\sin }^2\theta -\sin \theta =1$ and $0\le \theta \le \pi$, calculate the value of $\sin \theta$.

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $19$
• E. $30$

Answer 1

Answer: (C)

$\frac{1}{2}$

Given, $6{\sin }^2\theta -s\in \theta =1$

$\Rightarrow 6{\sin }^2\theta -\sin \theta -1=0$

$\Rightarrow 6{\sin }^2\theta -3\sin \theta +2\sin \theta -1=0$

$\Rightarrow 3\sin \theta (2\sin \theta -1)+1(2\sin \theta -1)=0$

$\Rightarrow (2\sin \theta -1)(3\sin \theta +1)=0$

If $2\sin \theta -1=0$

Then $\sin \theta =\frac{1}{2}$

If $3\sin \theta +1=0$

Then $\Rightarrow \sin \theta =-\frac{1}{3}$

The value of $0\le \theta \le \pi$

So, $\sin \theta =\frac{1}{2}$