Question

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

Answer 1

Finding the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathit{x}$ :

Given that

$$\begin{gathered} 5({\tan }^{2}x-{\cos }^{2}x)=2{\cos }^{2}x+9, \\ 5se{c}^{2}x–5=7{\cos }^{2}x+9 \end{gathered}$$

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

$$\begin{gathered} \Rightarrow \left(\frac{5}{t}\right)=9t+12 \\ \Rightarrow 9t^2+12t–5=0 \\ \Rightarrow t=\frac{1}{3}ast\ne –\frac{5}{3} \\ \Rightarrow {\cos }^{2}x=\frac{1}{3},(\cos 2x=2{\cos }^{2}x–1=-\frac{1}{3}) \\ \Rightarrow \cos 4x=2{\cos }^{2}2x–1 \\ =\left(\frac{2}{9}\right)–1 \\ =–\frac{7}{9} \end{gathered}$$

Hence, the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathit{x}$ is $-\frac{7}{9}$