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

Solution :
We have
$$\eqalign{ & 5\,{\tan ^2}x - 5\,{\cos ^2}x = 2\left( {{2\cos^2} 2x - 1} \right) + 9 \cr & \Rightarrow \,\,5\,{\tan ^2}x - 5\,{\cos ^2}x = 4\,{\cos ^2}x - 2 + 9 \cr & \Rightarrow \,\,5\,{\tan ^2}x = 9\,{\cos ^2}x + 7 \cr & \Rightarrow \,\,5\left( {{{\sec }^2}x - 1} \right) = 9\,{\cos ^2}x + 7 \cr & {\text{Let }}{\cos ^2}x = t \cr & \Rightarrow \,\,\frac{5}{t} - 9t - 12 = 0 \cr & \Rightarrow \,\,9{t^2} + 12t - 5 = 0 \cr & \Rightarrow \,\,9{t^2} + 15t - 3t - 5 = 0 \cr & \Rightarrow \,\,\left( {3t - 1} \right)\left( {3t + 5} \right) = 0 \cr & \Rightarrow \,\,t = \frac{1}{3}\,\,{\text{as }}\,t \ne - \frac{5}{3}. \cr & cos2x = 2co{s^2}x - 1 = 2\left( {\frac{1}{3}} \right) - 1 \cr & = - \frac{1}{3} \cr & \cos 4x = 2{\cos ^2}2x - 1 = 2{\left( { - \frac{1}{3}} \right)^2} - 1 \cr & = - \frac{7}{9} \cr} $$