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Question

Let $$\alpha$$ be the root of the equation $$25\cos^{2}\theta + 5\cos \theta - 12 = C$$, where $$\dfrac {\pi}{2} < \alpha < \pi$$.
What is $$\sin 2\alpha$$ equal to?


A
2425
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B
2425
loader
C
512
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D
2125
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Solution

The correct option is D $$\dfrac{-24}{25}$$
Consider that $$\cos\theta=x$$

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

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

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

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

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

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

$$\Rightarrow \sin2\alpha=2\sin\alpha\cos\alpha=2\times-\dfrac{4}{5}\times\dfrac{3}{5}=-\dfrac{24}{25}$$

This is the required solution.

Mathematics

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