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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$-\frac{7}{9}$

Vedclass · Answer

(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 $t = \cos^2 x$:$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

Similar Questions

If $f(\theta) = \cos^3 \theta + \cos^3 \left(\frac{2\pi}{3} + \theta\right) + \cos^3 \left(\theta - \frac{2\pi}{3}\right)$,then $f\left(\frac{\pi}{5}\right) = $

If angle $\theta$ is divided into two parts such that the tangent of one part is $k$ times the tangent of the other and $\phi$ is their difference,then $\sin \theta = $

If $\cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ} = k$,then $\frac{4k}{3} =$

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is equal to

If $\frac{1}{6} \sin \theta, \cos \theta$ and $\tan \theta$ are in geometric progression,then the solution set of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module