If 7 \sin^{4}\theta+9 \cos^{4}\theta+42 \sin^{2}\theta=16, 0 \lt \theta \lt \frac{\pi}{2} then what is \tan \theta equal to?

Explanation

Let t = \sin^2\theta. Then \cos^2\theta = 1-t. The equation becomes 7t^2 + 9(1-t)^2 + 42t = 16. Expanding: 7t^2 + 9(1-2t+t^2) + 42t = 16 \implies 16t^2 + 24t - 7 = 0. Using the quadratic formula, t = \frac{-24 \pm \sqrt{576 + 448}}{32} = \frac{-24 \pm 32}{32}. Since t > 0, t = \frac{8}{32} = \frac{1}{4}. Thus \sin^2\theta = \frac{1}{4}, yielding \sin\theta = \frac{1}{2}. Therefore \theta = 30^\circ and \tan\theta = \frac{1}{\sqrt{3}}.

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