Question Number 130819 by EDWIN88 last updated on 29/Jan/21 | ||
$$\:\mathrm{8sin}\:^{\mathrm{3}} \left({x}+\frac{\pi}{\mathrm{6}}\right)=\:\mathrm{cos}\:\left(\mathrm{3}{x}\right) \\ $$$$\:{x}=? \\ $$ | ||
Answered by mr W last updated on 29/Jan/21 | ||
$${let}\:{u}={x}+\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{3}{x}=\mathrm{3}{u}−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left(\mathrm{3}{x}\right)=\mathrm{cos}\:\left(\mathrm{3}{u}−\frac{\pi}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\mathrm{3}{u}\right) \\ $$$$=\mathrm{3}\:\mathrm{sin}\:{u}−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:{u} \\ $$$$\mathrm{8}\:\mathrm{sin}^{\mathrm{3}} \:{u}=\mathrm{3}\:\mathrm{sin}\:{u}−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:{u} \\ $$$$\left(\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{u}−\mathrm{1}\right)\mathrm{sin}\:{u}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:{u}=\mathrm{0}\:\Rightarrow{u}={k}\pi\:\Rightarrow{x}={k}\pi−\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{u}−\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{sin}\:{u}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{u}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{3}}\:\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{3}}\pm\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{u}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{3}}\:\Rightarrow{x}=\mathrm{2}{k}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}\pm\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$$${summary}: \\ $$$${x}={k}\pi−\frac{\pi}{\mathrm{6}} \\ $$$${x}={k}\pi \\ $$$${x}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{3}} \\ $$$${x}=\mathrm{2}{k}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ | ||
Commented by EDWIN88 last updated on 29/Jan/21 | ||
$${nice} \\ $$ | ||