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sinx-sin3x-sin-x-0-then-find-the-general-solution-




Question Number 27843 by bmind4860 last updated on 15/Jan/18
sinx+sin3x+sin(√x)=0, then find the   general solution?
$${sinx}+{sin}\mathrm{3}{x}+{sin}\sqrt{{x}}=\mathrm{0},\:{then}\:{find}\:{the}\: \\ $$$${general}\:{solution}? \\ $$
Commented by abdo imad last updated on 17/Jan/18
let put (√x)=t   e⇔  sint + sin(t^2 )  +sin(3t^2 ) general case  sint  + Σ_(k=0) ^n sin(2k+1)t^2  =0 also we can get geral case  Σ_(k=) ^n sin(2k+1)x +sin((√x))=0 for example let finf  Σ_(k=0) ^n sin(2k+1)t^2 =Σ_(k=0) ^n sin(2k+1)u  with u=t^2   and   Σ_(k=0) ^n  sin(2k+1)u=Im(Σ_(k=0 ) ^n   e^(i(2k+1)u) )  =Im(  e^(iu) Σ_(k=0) ^n  (e^(i(2u)) )^k ) = Im(  e^(iu)  ((1−(e^(i(2u)) )^(n+1) )/(1−e^(i2u) )))if  e^(i(2u)) ≠1   ......
$${let}\:{put}\:\sqrt{{x}}={t}\:\:\:{e}\Leftrightarrow\:\:{sint}\:+\:{sin}\left({t}^{\mathrm{2}} \right)\:\:+{sin}\left(\mathrm{3}{t}^{\mathrm{2}} \right)\:{general}\:{case} \\ $$$${sint}\:\:+\:\sum_{{k}=\mathrm{0}} ^{{n}} {sin}\left(\mathrm{2}{k}+\mathrm{1}\right){t}^{\mathrm{2}} \:=\mathrm{0}\:{also}\:{we}\:{can}\:{get}\:{geral}\:{case} \\ $$$$\sum_{{k}=} ^{{n}} {sin}\left(\mathrm{2}{k}+\mathrm{1}\right){x}\:+{sin}\left(\sqrt{{x}}\right)=\mathrm{0}\:{for}\:{example}\:{let}\:{finf} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} {sin}\left(\mathrm{2}{k}+\mathrm{1}\right){t}^{\mathrm{2}} =\sum_{{k}=\mathrm{0}} ^{{n}} {sin}\left(\mathrm{2}{k}+\mathrm{1}\right){u}\:\:{with}\:{u}={t}^{\mathrm{2}} \\ $$$${and}\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left(\mathrm{2}{k}+\mathrm{1}\right){u}={Im}\left(\sum_{{k}=\mathrm{0}\:} ^{{n}} \:\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right){u}} \right) \\ $$$$={Im}\left(\:\:{e}^{{iu}} \sum_{{k}=\mathrm{0}} ^{{n}} \:\left({e}^{{i}\left(\mathrm{2}{u}\right)} \right)^{{k}} \right)\:=\:{Im}\left(\:\:{e}^{{iu}} \:\frac{\mathrm{1}−\left({e}^{{i}\left(\mathrm{2}{u}\right)} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{{i}\mathrm{2}{u}} }\right){if} \\ $$$${e}^{{i}\left(\mathrm{2}{u}\right)} \neq\mathrm{1}\:\:\:…… \\ $$

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