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Question Number 206466 by cortano21 last updated on 15/Apr/24

Commented by cortano21 last updated on 15/Apr/24

Answered by mr W last updated on 15/Apr/24

$$\tan 3\alpha =2\tan 2\alpha$$$$\frac{\tan \alpha (3-{\tan }^2\alpha )}{1-3{\tan }^2\alpha }=\frac{4\tan \alpha }{1-{\tan }^2\alpha }$$$${\tan }^4\alpha +8{\tan }^2\alpha -1=0$$$${\tan }^2\alpha =\sqrt{17}-4$$$$x=12\sin 2\alpha \tan 2\alpha =\frac{48{\tan }^2\alpha }{1-{\tan }^4\alpha }$$$$=\frac{48(\sqrt{17}-4)}{1-{(\sqrt{17}-4)}^2}=6$$

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