Theta Cosine Sum Evaluation
If $ 0 < \theta < \frac{\pi}{2} $ and $ \sqrt{3} \cos \theta - \sin \theta = \frac{1}{3}, $ then find $ \sqrt{3} \sin \theta + \cos \theta $.
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Solution
From $ \sqrt{3} \cos \theta - \sin \theta = \frac{1}{3}, $
\[\sin \theta = \sqrt{3} \cos \theta - \frac{1}{3}.\]Substituting into $ \sin^2 \theta + \cos^2 \theta = 1, $ we get
\[3 \cos^2 \theta - \frac{2 \sqrt{3}}{3} \cos \theta + \frac{1}{9} + \cos^2 \theta = 1.\]This simplifies to $ 18 \cos^2 \theta - 3 \sqrt{3} \cos \theta - 4 = 0 $. By the quadratic formula,
\[\cos \theta = \frac{\sqrt{3} \pm \sqrt{35}}{12}.\]Since $ 0 < \theta < \frac{\pi}{2}, $ $ \cos \theta $ is positive, so $ \cos \theta = \frac{\sqrt{3} + \sqrt{35}}{12} $.
Hence,
\begin{align*}
\sqrt{3} \sin \theta + \cos \theta &= \sqrt{3} \left( \sqrt{3} \cos \theta - \frac{1}{3} \right) + \cos \theta \\
&= 3 \cos \theta - \frac{\sqrt{3}}{3} + \cos \theta \\
&= 4 \cos \theta - \frac{\sqrt{3}}{3} \\
&= \frac{\sqrt{3} + \sqrt{35}}{3} - \frac{\sqrt{3}}{3} \\
&= \boxed{\frac{\sqrt{35}}{3}}.\end{align*}
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