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Question Number 12532 by tawa last updated on 24/Apr/17

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:\mathrm{p}$$$${\tan }^{-1}\left(2\mathrm{x}\right)+{\tan }^{-1}\left(3\mathrm{x}\right)=\frac{\pi }{4}$$

Answered by mrW1 last updated on 25/Apr/17

$$\tan [{\tan }^{-1}\left(2\mathrm{x}\right)+{\tan }^{-1}\left(3\mathrm{x}\right)]=\tan \left(\frac{\pi }{4}\right)$$$$\frac{2x+3x}{1-2x\times 3x}=1$$$$5x=1-6x^2$$$$6x^2+5x-1=0$$$$(6x-1)(x+1)=0$$$$x=\frac{1}{6}$$$$x=-1(nosolution,sincex>0)$$

Commented by tawa last updated on 25/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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