Question

lf $\alpha$ is a root of $25{\cos }^2\theta +5\cos \theta -12=0,\frac{\pi }{2}<\alpha <p$, then the value of $\sin 2\alpha$ is equal to :

• A. $\frac{24}{25}$
• B. $-\frac{24}{25}$
• C. $\frac{13}{18}$
• D. $-\frac{13}{18}$

Answer 1

The correct option is B $-\frac{24}{25}$

$25{\cos }^2\theta +5\cos \theta -12=0\Rightarrow \cos \theta =\cos \alpha =\frac{-4}{5},\frac{3}{5}$

Since, $\frac{\pi }{2}<\alpha <\pi$ $\cos \alpha =\frac{-4}{5}\sin \alpha =\sqrt{1-(\frac{-4}{5}{)}^2}=\frac{3}{5}$

Now, $\sin 2\alpha =2\sin \alpha \cos \alpha =2\ast \frac{3}{5}\ast \frac{-4}{5}=\frac{-24}{25}$