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

$\begin{array}{1 1} (1) \large\frac{1}{3} \\ (2)\large\frac{2}{9} \\ (3) \large\frac{-7}{9} \\ (4) \large\frac{-3}{5} \end{array} $

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