We know that \[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \] Given that \[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1}{3} \Rightarrow \tan^2 \theta = \frac{1}{2} \] Putting this value in the equation, we have: \[ 32 \left( \frac{1}{2} \right)^4 = 2 \cos^2 \alpha - 3 \cos \alpha \] \[ \Rightarrow (2 \cos \alpha + 1)(\cos \alpha - 2) = 0 \] Or \[ \cos \alpha = - \frac{1}{2} \Rightarrow \alpha = 2n\pi \pm \frac{2\pi}{3} \]