If $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$,then $\cos 4x$ is equal to

  • A
    $-\frac{7}{9}$
  • B
    $-\frac{3}{5}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{2}{9}$

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