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