Given that, $5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$ $\Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9$ Let ${\cos ^2}x = t,$ then we have $5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9$ $\Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9$ $\Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$ $\Rightarrow 9{t^2} + 12t - 5 = 0$ $\Rightarrow 9{t^2} + 15t - 3t - 5 = 0$ $\Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0$ $\Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0$ $\therefore$ $t = {1 \over 3}$ and $t = - {5 \over 3}$ If $t = - {5 \over 3}$ then ${\cos ^2}x$ is negative So , $t$ can not be $- {5 \over 3}$. So, correct value of $t = {1 \over 3}$ then $\cos {}^2x = t = {1 \over 3}$ $\therefore\,\,\,$ $\cos 4x$ $= 2{\cos ^2}2x - 1$ $= 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1$ $= 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1$ $= 2.{\left( { - {1 \over 3}} \right)^2} - 1$ $= {2 \over 9} - 1$ $= - {7 \over 9}$