If alpha is a root of 25{cos ^2}theta + 5cos | vedclass

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Question

(b) Since $\alpha$ is a root of $25{\cos ^2}	heta + 5\cos 	heta - 12 = 0$ 

$	herefore$ $25{\cos ^2}\alpha + 5\cos \alpha - 12 = 0$&nbs

Vedclass · Accepted Answer

$ - 24/25$

Vedclass · Answer

(b) Since $\alpha$ is a root of $25{\cos ^2} heta + 5\cos heta - 12 = 0$ $ herefore$ $25{\cos ^2}\alpha + 5\cos \alpha - 12 = 0$ ==> $\cos \alpha = \frac{{ - 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ - 5 \pm 35}}{{50}}$ ==> $\cos \alpha = - 4/5$ $[ \because \pi /2 < \alpha < \pi \Rightarrow \cos \alpha < 0]$ $ herefore $$\sin 2\alpha = 2\sin \alpha \cos \alpha = - 24/25$.

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

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