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Question-183076




Question Number 183076 by Engr_Jidda last updated on 19/Dec/22
Answered by TheSupreme last updated on 21/Dec/22
t = tan(x)  10≥t≥(7/3)  (t−2)^π +(t+5)^π =(3t−7)^π +(10−t)^π   t−2=10−t ∨ t+5=3t−7  t=6 ∨ t=6 ✓ t=6 is a solution  t−2=3t−7 ∨ 10−t = t+5  t=(5/2) ∨ t=(5/2) ✓ is a solution    we have to proove that there are only 2 solutions
$${t}\:=\:{tan}\left({x}\right) \\ $$$$\mathrm{10}\geqslant{t}\geqslant\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\left({t}−\mathrm{2}\right)^{\pi} +\left({t}+\mathrm{5}\right)^{\pi} =\left(\mathrm{3}{t}−\mathrm{7}\right)^{\pi} +\left(\mathrm{10}−{t}\right)^{\pi} \\ $$$${t}−\mathrm{2}=\mathrm{10}−{t}\:\vee\:{t}+\mathrm{5}=\mathrm{3}{t}−\mathrm{7} \\ $$$${t}=\mathrm{6}\:\vee\:{t}=\mathrm{6}\:\checkmark\:{t}=\mathrm{6}\:{is}\:{a}\:{solution} \\ $$$${t}−\mathrm{2}=\mathrm{3}{t}−\mathrm{7}\:\vee\:\mathrm{10}−{t}\:=\:{t}+\mathrm{5} \\ $$$${t}=\frac{\mathrm{5}}{\mathrm{2}}\:\vee\:{t}=\frac{\mathrm{5}}{\mathrm{2}}\:\checkmark\:{is}\:{a}\:{solution} \\ $$$$ \\ $$$${we}\:{have}\:{to}\:{proove}\:{that}\:{there}\:{are}\:{only}\:\mathrm{2}\:{solutions} \\ $$

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