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Equations in sin²θ, cos²θ, tan²θ

Trigonometric equations

Concept 4 of 15 in Trigonometric Functions

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Ravi solves tan⁡2θ=13\tan^2\theta = \frac{1}{3} as tan⁡θ=13\tan\theta = \frac{1}{\sqrt{3}}, so θ=nπ+π6\theta = n\pi + \frac{\pi}{6}. Is θ=5π6\theta = \frac{5\pi}{6} a solution he missed?