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Question-190938




Question Number 190938 by Shrinava last updated on 14/Apr/23
Answered by Frix last updated on 15/Apr/23
tan 3x −tan x tan 11x tan 13x =0  We can transform this to  sin 28x −sin 20x +sin 4x =0  y=4x  sin 7y −sin 5y +sin y =0  s=sin y & transforming  s(s^6 −((3s^4 )/2)+((9s^2 )/(16))−(3/(64)))=0  s=0 ★  t=s^2   t^3 −((3t^2 )/2)+((9t)/(16))−(3/(64))=0  u=t−(1/2)  u^3 −((3u)/(16))−(1/(64))=0  u=−((cos ((2π)/9))/2)∨u=−((sin (π/(18)))/2)∨u=((cos (π/9))/2) ★  Now we have to go back  For 0≤x<(π/2) I get  (x/π)∈{0, (1/(36)), (1/(18)), (1/9), (5/(36)), (7/(36)), (2/9), (1/4), (5/(18)), ((11)/(36)), ((13)/(36)), (7/(18)), (4/9), ((17)/(36))}
$$\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:{x}\:\mathrm{tan}\:\mathrm{11}{x}\:\mathrm{tan}\:\mathrm{13}{x}\:=\mathrm{0} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{transform}\:\mathrm{this}\:\mathrm{to} \\ $$$$\mathrm{sin}\:\mathrm{28}{x}\:−\mathrm{sin}\:\mathrm{20}{x}\:+\mathrm{sin}\:\mathrm{4}{x}\:=\mathrm{0} \\ $$$${y}=\mathrm{4}{x} \\ $$$$\mathrm{sin}\:\mathrm{7}{y}\:−\mathrm{sin}\:\mathrm{5}{y}\:+\mathrm{sin}\:{y}\:=\mathrm{0} \\ $$$${s}=\mathrm{sin}\:{y}\:\&\:\mathrm{transforming} \\ $$$${s}\left({s}^{\mathrm{6}} −\frac{\mathrm{3}{s}^{\mathrm{4}} }{\mathrm{2}}+\frac{\mathrm{9}{s}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{64}}\right)=\mathrm{0} \\ $$$${s}=\mathrm{0}\:\bigstar \\ $$$${t}={s}^{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{9}{t}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{64}}=\mathrm{0} \\ $$$${u}={t}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{3}{u}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{64}}=\mathrm{0} \\ $$$${u}=−\frac{\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{2}}\vee{u}=−\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{18}}}{\mathrm{2}}\vee{u}=\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{9}}}{\mathrm{2}}\:\bigstar \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{go}\:\mathrm{back} \\ $$$$\mathrm{For}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}}\:\mathrm{I}\:\mathrm{get} \\ $$$$\frac{{x}}{\pi}\in\left\{\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{36}},\:\frac{\mathrm{1}}{\mathrm{18}},\:\frac{\mathrm{1}}{\mathrm{9}},\:\frac{\mathrm{5}}{\mathrm{36}},\:\frac{\mathrm{7}}{\mathrm{36}},\:\frac{\mathrm{2}}{\mathrm{9}},\:\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{5}}{\mathrm{18}},\:\frac{\mathrm{11}}{\mathrm{36}},\:\frac{\mathrm{13}}{\mathrm{36}},\:\frac{\mathrm{7}}{\mathrm{18}},\:\frac{\mathrm{4}}{\mathrm{9}},\:\frac{\mathrm{17}}{\mathrm{36}}\right\} \\ $$

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