Question

If $5({\tan }^{2}x-{\cos }^{2}x)=2\cos x+9$, then the value of $\cos 4x$ is.

Answer 1

We have,

$5\left({\sec }^{2}x-1-{\cos }^{2}x\right)=2\left(2{\cos }^{2}x-1\right)+9$

Let

${\cos }^{2}x=t$

$5\left(\frac{1}{t}-1-t\right)=2\left(2t-1\right)+95\left(1-t-t^2\right)=4t^2+7t9t^2+12t-5=0t=\frac{1}{3}.\frac{-5}{3}{\cos }^{2}x=\frac{1}{3}\cos 2x=\frac{2}{3}-1=\frac{-1}{3}\cos 4x=\frac{-7}{9}$

Hence, this is the answer.