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)

$-\frac{7}{9}$

$5({\tan }^{2}x-{\cos }^{2}x)=2\cos 2x+9$

$\Rightarrow 5(\frac{{\sin }^{2}x}{{\cos }^{2}x}-{\cos }^{2}x)=2(2{\cos }^{2}x-1)+9$

$\Rightarrow 5\frac{({\sin }^{2}x-{\cos }^{4}x)}{{\cos }^{2}x}=4{\cos }^{2}x-2+9$

$\Rightarrow 5(1-{\cos }^{2}x-{\cos }^{4}x)=4{\cos }^{4}x+7{\cos }^{2}x$

$\Rightarrow 5-5{\cos }^{2}x-5{\cos }^{4}x=4{\cos }^{4}x+7{\cos }^{2}x$

$\Rightarrow 9{\cos }^{4}x+12{\cos }^{2}x-5=0$

$\Rightarrow 9({\cos }^{2}x{)}^2+12{\cos }^{2}x-5=0$

Using quadratic formula,

${\cos }^{2}x=\frac{-12+\sqrt{(12{)}^2+4(5\left)\right(9)}}{2\times 9}$

$=\frac{-12\pm \sqrt{144+180}}{18}$

$=\frac{-12\pm \sqrt{324}}{18}$

$=\frac{-12\pm 18}{18}$

$\Rightarrow {\cos }^{2}x=\frac{-12+18}{18}$ or ${\cos }^{2}x=\frac{-12-18}{18}$

$=\frac{6}{18}$ $=\frac{-30}{18}=\frac{-5}{3}$, which is neglected

$=\frac{1}{3}$

$\therefore \cos 4c=2{\cos }^{2}2x-1$

$=2(2{\cos }^{2}x-1{)}^2-1$

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

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

$=2\times {\left(\frac{-1}{3}\right)}^2-1=\frac{2}{9}-1=\frac{-7}{9}$.