If 5(tan^2 x - cos^2 x) = 2cos 2x + 9,then co | vedclass

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(A) Given equation: $5(	an^2 x - \cos^2 x) = 2\cos 2x + 9$Using $\cos 2x = 2\cos^2 x - 1$,we get:$5	an^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$$5	an

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$-\frac{7}{9}$

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(A) Given equation: $5(	an^2 x - \cos^2 x) = 2\cos 2x + 9$Using $\cos 2x = 2\cos^2 x - 1$,we get:$5	an^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$$5	an^2 x - 5\cos^2 x = 4\cos^2 x - 2 + 9$$5	an^2 x = 9\cos^2 x + 7$Since $	an^2 x = \sec^2 x - 1 = \frac{1}{\cos^2 x} - 1$,let $\cos^2 x = t$:$5(\frac{1}{t} - 1) = 9t + 7$$5 - 5t = 9t^2 + 7t$$9t^2 + 12t - 5 = 0$$(3t - 1)(3t + 5) = 0$Since $t = \cos^2 x$ must be positive,$t = \frac{1}{3}$.Now,$\cos 2x = 2\cos^2 x - 1 = 2(\frac{1}{3}) - 1 = -\frac{1}{3}$.Finally,$\cos 4x = 2\cos^2 2x - 1 = 2(-\frac{1}{3})^2 - 1 = 2(\frac{1}{9}) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

If $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$,then $\cos 4x$ is equal to

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