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8sin-3-x-pi-6-cos-3x-x-




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

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