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 $25{\cos }^2\theta +5\cos \theta -12=0$
$\therefore 25{\cos }^2\alpha +5\cos \alpha -12=0$
$\Rightarrow 25{\cos }^2\alpha +20\cos \alpha -15\cos ϵ-12=0$
$\Rightarrow 5\cos \alpha (5\cos \alpha +4)-3(5\cos \alpha +4)=0$
$\Rightarrow \cos \alpha =-\frac{4}{5},\frac{3}{5}$
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}$
$=-\frac{24}{25}$