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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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