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

The correct option is B

$-\frac{24}{25}$

Explanation for the correct option:

Step 1. Find the value of $\sin 2\alpha$:

Given, $\alpha$ is a root of $25{\cos }^2\theta +5\cos \theta –12=0$.

$\therefore 25{\cos }^2\alpha +5\cos \alpha –12=0$

$\Rightarrow$ $25{\cos }^2\alpha +20\cos \alpha –15\cos \alpha –12=0$

$\Rightarrow$ $5\cos \alpha (5\cos \alpha +4)–3(5\cos \alpha +4)=0$

$\Rightarrow$ $(5\cos \alpha –3)(5\cos \alpha +4)=0$

$\Rightarrow$$\cos \alpha =\frac{3}{5},\cos \alpha =-\frac{4}{5}$

Step 2. Since $\alpha$ lies in the quadrant II and $\cos \alpha$ is negative in this quadrant, we get

$\Rightarrow$$\cos \alpha =-\frac{4}{5}$

$\Rightarrow$ $\sin \alpha =\frac{3}{5}$

$\begin{array}{rcl}\therefore \sin 2\alpha & =& 2\sin \alpha \cos \alpha \\ & =& 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)\\ & =& \mathbf{-}\frac{\mathbf{24}}{\mathbf{25}}\end{array}$

Hence, Option ‘B’ is Correct.