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I'm bad at formatting. Sorry for that. Using half angles, cos 2@ is (1-tan^2@)/(1+tan^2@) . Solve it n get tan @ as 0.5 n cos 2@ as 0.6.

Who imagined 1/2 and 3/5 share such a good relation?

$\cos {2\theta} = \frac{6}{5} \tan {\theta}\quad \Rightarrow 2\cos^2{\theta} - 1 = \frac {6}{5} \tan {\theta}\quad \Rightarrow \dfrac {10}{\sec^2{\theta} } - 5 = 6\tan {\theta}$

$\Rightarrow \dfrac {10}{1+\tan^2{\theta}} - 5 = 6 \tan {\theta}\quad \Rightarrow 10- 5 - 5\tan^2{\theta} = 6\tan {\theta} +6 \tan^3{\theta}$

$\Rightarrow 6 \tan^3{\theta} + 5\tan^2{\theta} + 6\tan {\theta} - 5 = 0$

$\Rightarrow (2 \tan {\theta} - 1)(3 \tan^2 {\theta} + 4 \tan {\theta} + 5) = 0 \quad \Rightarrow \tan {\theta} = \boxed{\frac{1}{2}}$