If alpha is a root of the equation 25 cos^2 t | vedclass

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(B) Given the equation $25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0$. Let $x = \cos \alpha$. Then $25x^2 + 5x - 12 = 0$. Using the quadratic formula $x

Vedclass · Accepted Answer

$\frac{-24}{25}$

Vedclass · Answer

(B) Given the equation $25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0$. Let $x = \cos \alpha$. Then $25x^2 + 5x - 12 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-5 \pm \sqrt{25 - 4(25)(-12)}}{50} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm 35}{50}$. So,$x = \frac{30}{50} = \frac{3}{5}$ or $x = \frac{-40}{50} = -\frac{4}{5}$. Since $\frac{\pi}{2} Therefore,$\cos \alpha = -\frac{4}{5}$. Now,$\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ (since $\sin \alpha > 0$ in the second quadrant). Finally,$\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) = -\frac{24}{25}$.

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

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