If sin^2 x + sin x cos x - 6cos^2 x = 0 and - | vedclass

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Question

(C) Divide the equation $\sin^2 x + \sin x \cos x - 6\cos^2 x = 0$ by $\cos^2 x$ (since $\cos x 
eq 0$ for $-\frac{\pi}{2} < x < 0$): $	an^2 x + 	a

Vedclass · Accepted Answer

$-\frac{4}{5}$

Vedclass · Answer

(C) Divide the equation $\sin^2 x + \sin x \cos x - 6\cos^2 x = 0$ by $\cos^2 x$ (since $\cos x 
eq 0$ for $-\frac{\pi}{2} $	an^2 x + 	an x - 6 = 0$ Factor the quadratic equation: $(	an x + 3)(	an x - 2) = 0$ This gives $	an x = -3$ or $	an x = 2$. Since $-\frac{\pi}{2} Using the formula $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$: $\cos 2x = \frac{1 - (-3)^2}{1 + (-3)^2} = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -\frac{4}{5}$.

If $\sin^2 x + \sin x \cos x - 6\cos^2 x = 0$ and $-\frac{\pi}{2} < x < 0$,then the value of $\cos 2x$ is

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