Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{9}$
• C. $-\frac{7}{9}$
• D. $-\frac{3}{5}$

Answer 1

Answer: (C)

$5{\tan }^{2}x=9{\cos }^{2}x+7$
$5{\sec }^{2}x-5=9{\cos }^{2}x+7$
Let ${\cos }^{2}x=t$
$\frac{5}{t}=9t+12$
$9t^2+12t-5=0$
$t=\frac{1}{3}$ as $t\ne -\frac{5}{3}$
${\cos }^{2}x=\frac{1}{3},\cos 2x=2{\cos }^{2}x-1$
$=-\frac{1}{3}$
$\cos 4x=2{\cos }^{2}2x-1$
$=\frac{2}{9}-1$
$=-\frac{7}{9}$