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Question Number 106579 by john santu last updated on 06/Aug/20

$$\mathcal{G}\mathrm{iven}\cos \mathrm{x}+\cos \mathrm{y}=\frac{3}{2}+\cos (\mathrm{x}+\mathrm{y})$$ $$\mathrm{where}\mathrm{x},\mathrm{y}\in [0,2\pi ].\mathrm{find}\mathrm{x}\mathrm{\&}\mathrm{y}$$

Answered by bobhans last updated on 06/Aug/20

$$\cos \mathrm{x}+\cos \mathrm{y}=\frac{3}{2}+\cos \mathrm{xcos}\mathrm{y}-\sin \mathrm{xsin}\mathrm{y}$$ $$\cos \mathrm{x}+\cos \mathrm{y}-\cos \mathrm{xcos}\mathrm{y}+\sin \mathrm{xsin}\mathrm{y}=\frac{3}{2}$$ $$\mathrm{recall}\to \{\begin{array}{l}\cos \mathrm{x}=\frac{1-\tan {}^2\left(\frac{\mathrm{x}}{2}\right)}{1+\tan {}^2\left(\frac{\mathrm{x}}{2}\right)}\\ \sin \mathrm{x}=\frac{2\tan \left(\frac{\mathrm{x}}{2}\right)}{1+\tan {}^2\left(\frac{\mathrm{x}}{2}\right)}\end{array}$$ $$\mathrm{set}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{p}\mathrm{\&}\tan \left(\frac{\mathrm{y}}{2}\right)=\mathrm{q}$$ $$\iff \frac{1-{\mathrm{p}}^2}{1+{\mathrm{p}}^2}+\frac{1-{\mathrm{q}}^2}{1+{\mathrm{q}}^2}-\left(\frac{1-{\mathrm{p}}^2}{1+{\mathrm{p}}^2}\right)\left(\frac{1-{\mathrm{q}}^2}{1+{\mathrm{q}}^2}\right)+\frac{4\mathrm{pq}}{(1+{\mathrm{p}}^2)(1+{\mathrm{q}}^2)}=\frac{3}{2}$$ $${9\mathrm{p}}^2{\mathrm{q}}^2-8\mathrm{pq}+{\mathrm{p}}^2+{\mathrm{q}}^2+1=0$$ $${(3\mathrm{pq}-1)}^2+{(\mathrm{p}-\mathrm{q})}^2=0$$ $$\to \{\begin{array}{l}\mathrm{p}=\mathrm{q}\\ \mathrm{p}=\pm \frac{1}{\sqrt{3}}\end{array}$$ $$\mathrm{case}\left(1\right)\{\begin{array}{l}\mathrm{p}=\frac{1}{\sqrt{3}}\Rightarrow \tan \left(\frac{\mathrm{x}}{2}\right)=\tan \left(\frac{\pi }{6}\right)\\ \mathrm{q}=\frac{1}{\sqrt{3}}\Rightarrow \tan \left(\frac{\mathrm{y}}{2}\right)=\tan \left(\frac{\pi }{6}\right)\end{array}$$ $$\to \{\begin{array}{l}\mathrm{x}=\frac{\pi }{3}\\ \mathrm{y}=\frac{\pi }{3}\end{array}\Rightarrow (\frac{\pi }{3},\frac{\pi }{3})$$ $$\mathrm{case}\left(2\right)$$ $$\to \{\begin{array}{l}\mathrm{p}=-\frac{1}{\sqrt{3}}\Rightarrow \tan \left(\frac{\mathrm{x}}{2}\right)=\tan (-\frac{\pi }{6})\\ \mathrm{q}=-\frac{1}{\sqrt{3}}\Rightarrow \tan \left(\frac{\mathrm{y}}{2}\right)=\tan (-\frac{\pi }{6})\end{array}$$ $$\to \{\begin{array}{l}\mathrm{x}=\frac{5\pi }{3}\\ \mathrm{y}=\frac{5\pi }{3}\end{array}\Rightarrow (\frac{5\pi }{3},\frac{5\pi }{3})$$

Commented byjohn santu last updated on 06/Aug/20

$$\mathrm{cool}...\mathrm{\&}\mathrm{jooss}\mathrm{♠}$$

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