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sin-3x-1-x-cos-3x-cos-x-2-




Question Number 193668 by SAMIRA last updated on 17/Jun/23
((sin(3x))/(1−x))−((cos(3x))/(cos(x))) = −2  ????
$$\frac{\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{1}−\mathrm{x}}−\frac{\mathrm{cos}\left(\mathrm{3x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)}\:=\:−\mathrm{2}\:\:????\:\: \\ $$
Answered by Frix last updated on 18/Jun/23
Transform to  x=−((4sin^3  x −4sin^2  x −3sin x −1)/(1+4sin^2  x))=f(x)  It′s easy to show that  1−((√(−9+6(√3)))/2)≤f(x)≤1+((√(−9+6(√3)))/2)  ≈.41≤f(x)≤1.59  ⇒  We find the solution for .41≤x≤1.59  ((sin 3x)/(1−x))−((cos 3x)/(cos x))+2=0  (((3−4sin^2  x)sin x)/(1−x))+1+4sin^2  x =0  We see there′s a change of sign around the  singularity at x=1 ⇒ we find the solution  for 1<x≤1.59  We can only approximate  ⇒  x≈1.02049
$$\mathrm{Transform}\:\mathrm{to} \\ $$$${x}=−\frac{\mathrm{4sin}^{\mathrm{3}} \:{x}\:−\mathrm{4sin}^{\mathrm{2}} \:{x}\:−\mathrm{3sin}\:{x}\:−\mathrm{1}}{\mathrm{1}+\mathrm{4sin}^{\mathrm{2}} \:{x}}={f}\left({x}\right) \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{1}−\frac{\sqrt{−\mathrm{9}+\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{2}}\leqslant{f}\left({x}\right)\leqslant\mathrm{1}+\frac{\sqrt{−\mathrm{9}+\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$\approx.\mathrm{41}\leqslant{f}\left({x}\right)\leqslant\mathrm{1}.\mathrm{59} \\ $$$$\Rightarrow \\ $$$$\mathrm{We}\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{for}\:.\mathrm{41}\leqslant{x}\leqslant\mathrm{1}.\mathrm{59} \\ $$$$\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{1}−{x}}−\frac{\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{cos}\:{x}}+\mathrm{2}=\mathrm{0} \\ $$$$\frac{\left(\mathrm{3}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)\mathrm{sin}\:{x}}{\mathrm{1}−{x}}+\mathrm{1}+\mathrm{4sin}^{\mathrm{2}} \:{x}\:=\mathrm{0} \\ $$$$\mathrm{We}\:\mathrm{see}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{around}\:\mathrm{the} \\ $$$$\mathrm{singularity}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{for}\:\mathrm{1}<{x}\leqslant\mathrm{1}.\mathrm{59} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\Rightarrow \\ $$$${x}\approx\mathrm{1}.\mathrm{02049} \\ $$

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