Question

Let $\alpha$ be the root of the equation $25{\cos }^2\theta +5\cos \theta -12=C$, where $\frac{\pi }{2}<\alpha <\pi$.

What is $\sin 2\alpha$ equal to?

• A. $\frac{24}{25}$
• B. $\frac{-24}{25}$
• C. $\frac{-5}{12}$
• D. $\frac{-21}{25}$

Answer 1

Answer: (B)

$\frac{-24}{25}$

Consider that $\cos \theta =x$

$\Rightarrow 25x^2+5x-12=0$

$\Rightarrow 25x^2+20x-15x-12=0$

$\Rightarrow 5x(5x+4)-3(5x+4)=0$

$\Rightarrow (5x-3)(5x+4)=0$

$\Rightarrow x=-\frac{4}{5}$ or $\frac{3}{5}$

Now, as $\alpha \in (\frac{\pi }{2},\pi )$, we have $\cos \alpha =-\frac{4}{5}$

$\Rightarrow \sin 2\alpha =2\sin \alpha \cos \alpha =2\times -\frac{4}{5}\times \frac{3}{5}=-\frac{24}{25}$

This is the required solution.