Question

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

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

Answer 1

Answer: (B)

$\because \alpha$ is a root of $25 \cos ^{2} \theta+5 \cos \theta-12=0$
$\therefore \quad 25 \cos ^{2} \alpha+5 \cos \alpha-12=0$
$\Rightarrow 25 \cos ^{2} \alpha+20 \cos \alpha-15 \cos \epsilon-12=0$
$\Rightarrow 5 \cos \alpha(5 \cos \alpha+4)-3(5 \cos \alpha+4)=0$
$\Rightarrow \cos \alpha=-\frac{4}{5}, \frac{3}{5}$
But $\frac{\pi}{2} < \alpha<\pi$
$\cos \alpha=-\frac{4}{5} (\because \cos \alpha<0)$
$ \Rightarrow \sin \alpha= \frac{3}{5} $
$\therefore \sin 2 \alpha=2 \sin \alpha \cos \alpha $$=-2 \times \frac{3}{5} \times \frac{4}{5} $
$=-\frac{24}{25}$