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Question Number 29658 by gyugfeet last updated on 11/Feb/18

$$tan\theta +tan2\theta +\sqrt{3}tan\theta .tan2\theta =\sqrt{3}(0^o⩽\theta ⩽360)$$

Answered by mrW2 last updated on 11/Feb/18

$$tan\theta +tan2\theta =\sqrt{3}(1-tan\theta .tan2\theta )$$$$\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=\sqrt{3}$$$$\tan 3\theta =\sqrt{3}$$$$3\theta =n\pi +\frac{\pi }{3}$$$$\theta =\frac{n\pi }{3}+\frac{\pi }{9}$$$$within[0,2\pi ]:$$$$\theta =\frac{\pi }{9},\frac{4\pi }{9},\frac{7\pi }{9},\frac{10\pi }{9},\frac{13\pi }{9},\frac{16\pi }{9}$$$$or$$$$\theta =20°,80°,140°,200°,260°,320°.$$

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