If $7 cos2 \theta+ 3 sin2 \theta= 4 and 0 < \theta< \frac {\pi}{2} $, then what is the value of $ \tan \theta$?
Solution
$7 \cos^2 \theta + 3 \sin \theta = 4$
$7 + \dfrac{3 \sin 2 \theta}{\cos 2 \theta} = \dfrac{4}{\cos2 \theta}$
$7 + 3 \tan 2 \theta = \dfrac{4}{\cos 2 \theta}$
$7 + 3 \tan \left[\dfrac{2 \tan \theta}{1 - \tan^2 \theta}\right] = \dfrac{4(1 + \tan^2 \theta)}{(1 - \tan^2 \theta)}$
$7 - 7\tan^2 \theta + 6 \tan \theta = 4 + 4 \tan^2 \theta$
$11 \tan^2 \theta - 6 \tan \theta - 3 = 0$
$\tan \theta = \sqrt{3}$