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Question Number 44172 by rahul 19 last updated on 22/Sep/18

$$Thenumberofsolutionsoftheequation$$$${\cos }^{-1}\frac{x^2-1}{x^2+1}+{\sin }^{-1}\frac{2x}{x^2+1}+{\tan }^{-1}\frac{2x}{x^2-1}=\frac{2\pi }{3}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Sep/18

Answered by MJS last updated on 22/Sep/18

$$x=\tan \frac{\theta }{2}$$$$\frac{x^2-1}{x^2+1}=-\cos \theta \frac{2x}{x^2+1}=\sin \theta \frac{2x}{x^2-1}=-\tan \theta$$$$\mathrm{but}\arccos (-\cos \theta ),\arcsin \left(\sin \theta \right)\mathrm{and}$$$$\arctan (-\tan \theta )\mathrm{are}\mathrm{strange}\mathrm{functions}\mathrm{and}$$$$\mathrm{we}\mathrm{get}(z\in \mathbb{Z})$$$${\theta }_1=\frac{\pi }{3}+2\pi z\Rightarrow x_1=\frac{\sqrt{3}}{3}$$$${\theta }_2=\frac{7\pi }{9}+2\pi z\Rightarrow x_2=\cot \frac{\pi }{9}=\tan \frac{7\pi }{18}$$$${\theta }_3=\frac{5\pi }{3}+2\pi z\Rightarrow x_3=-\frac{\sqrt{3}}{3}$$

Commented by rahul 19 last updated on 23/Sep/18

$$Strangefunctionsmeans?$$$$Sir,ans.is2!$$

Commented by MJS last updated on 23/Sep/18

$$[0;2\pi [$$$$\arccos (-\cos \theta )=\{\begin{array}{l}\pi -\theta \wedge \theta \in [0;\pi [\\ \theta -\pi \wedge \theta \in [\pi ;2\pi [\end{array}$$$$\arcsin \left(\sin \theta \right)=\{\begin{array}{l}\theta \wedge \theta \in [0;\frac{\pi }{2}[\\ \pi -\theta \wedge \theta \in [\frac{\pi }{2};\frac{3\pi }{2}[\\ \theta -2\pi \wedge \theta \in [\frac{3\pi }{2};2\pi [\end{array}$$$$\arctan (-\tan \theta )=\{\begin{array}{l}-\theta \wedge \theta \in [0;\frac{\pi }{2}[\\ \pi -\theta \wedge \theta \in [\frac{\pi }{2};\frac{3\pi }{2}[\\ 2\pi -\theta \wedge \theta \in [\frac{3\pi }{2};2\pi [\end{array}$$$$\arccos (-\cos \theta )+\arcsin \left(\sin \theta \right)+\arctan (-\tan \theta )=$$$$=\{\begin{array}{l}\pi -\theta \wedge \theta \in [0;\frac{\pi }{2}[\\ 3(\pi -\theta )\wedge \theta \in [\frac{\pi }{2};\pi [\\ \pi -\theta \wedge \theta \in [\pi ;\frac{3\pi }{2}[\\ \theta -\pi \wedge \theta \in [\frac{3\pi }{2};2\pi [\end{array}$$$$\frac{2\pi }{3}=\pi -\theta \Rightarrow \theta =\frac{\pi }{3}\Rightarrow x=\frac{\sqrt{3}}{3}$$$$\frac{2\pi }{3}=3(\pi -\theta )\Rightarrow \theta =\frac{7\pi }{9}\Rightarrow x=\tan \frac{7\pi }{18}$$$$\frac{2\pi }{3}=\pi -\theta \Rightarrow \theta =\frac{\pi }{3}\notin [\pi ;\frac{3\pi }{2}[$$$$\frac{2\pi }{3}=\theta -\pi \Rightarrow \theta =\frac{5\pi }{3}\Rightarrow x=-\frac{\sqrt{3}}{3}$$

Commented by MJS last updated on 23/Sep/18

$$\mathrm{also}\mathrm{if}\mathrm{you}\mathrm{draw}\mathrm{the}\mathrm{original}\mathrm{equation}\mathrm{you}$$$$\mathrm{see}\mathrm{the}3\mathrm{solutions}$$

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