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

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

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$-24/25$

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(B) Given the quadratic equation $25\cos^2	heta + 5\cos	heta - 12 = 0$. Let $x = \cos	heta$,then $25x^2 + 5x - 12 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{-5 \pm \sqrt{25 - 4(25)(-12)}}{50} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm \sqrt{1225}}{50} = \frac{-5 \pm 35}{50}$.This gives two possible values for $\cos\alpha$: $x_1 = \frac{30}{50} = 3/5$ and $x_2 = \frac{-40}{50} = -4/5$.Since $\pi/2 Now,find $\sin\alpha$: $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (-4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$ (positive in the second quadrant).Finally,$\sin 2\alpha = 2\sin\alpha\cos\alpha = 2(3/5)(-4/5) = -24/25$.

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

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