Question

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

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

Answer 1

The correct option is D $-\frac{7}{9}$

We know that ${\tan }^{2}x={\sec }^{2}x-1$ and $\cos 2x=2{\cos }^{2}x-1$

$\therefore 5({\sec }^2-1-{\cos }^{2}x)=2[2{\cos }^{2}x-1]+9$

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

$\Rightarrow 5(\frac{1}{t}-1-t)=2(2t-1)+9$

$\Rightarrow \frac{5}{t}-5-5t=4t+7$

$\Rightarrow 5-5t-5t^2=4t^2+7t$

$\Rightarrow 9t^2+12t-5=0$

$\Rightarrow 9t^2+15t-3t-5=0$

$\Rightarrow (3t+5)(3t-1)=0$

$\Rightarrow t=-\frac{5}{3}$ or $t=\frac{1}{3}$

$t\ne -\frac{5}{3}$ as the range of ${\cos }^{2}x$ is $[0,1]$

Hence, ${\cos }^{2}x=\frac{1}{3}$

Now, $\cos 4x=2{\cos }^{2}2x-1=2[2{\cos }^{2}x-1{]}^2-1$

$=2{[2\left(\frac{1}{3}\right)-1]}^2-1$

$=\frac{2}{9}-1$

$=-\frac{7}{9}$

Option $D$ is the correct answer.