Solve-it-in-pi-pi-sin-2x-cos-x-

Question Number 75502 by mathocean1 last updated on 12/Dec/19

$$\left.\mathrm{S}\left.\mathrm{olve}\:\mathrm{it}\:\mathrm{in}\:\right]−\pi;\pi\right] \\ $$$$\mathrm{sin}\left(\mathrm{2x}\right)=\mathrm{cos}\left(\mathrm{x}\right) \\ $$

Commented by mathmax by abdo last updated on 12/Dec/19

$${sin}\left(\mathrm{2}{x}\right)={cosx}\:\Leftrightarrow\:{sin}\left(\mathrm{2}{x}\right)={sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\:\Leftrightarrow\mathrm{2}{x}=\frac{\pi}{\mathrm{2}}−{x}+\mathrm{2}{k}\pi\:{or} \\ $$$$\mathrm{2}{x}\:=\frac{\pi}{\mathrm{2}}+{x}\:+\mathrm{2}{k}\pi\:\left(\:{kfrom}\:{Z}\right)\:\Leftrightarrow\mathrm{3}{x}\:=\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:{or}\:{x}=\frac{\pi}{\mathrm{2}}+\:\mathrm{2}{k}\pi\:\Leftrightarrow \\ $$$${x}\:=\frac{\pi}{\mathrm{6}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\:{or}\:{x}=\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\:\left({k}\in\:{Z}\right) \\ $$

Commented by mathmax by abdo last updated on 12/Dec/19

$${case}\:\mathrm{1}\:\:−\pi\:<\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:<\pi\:\Rightarrow−\mathrm{1}<\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{2}{k}<\mathrm{1}\:\Rightarrow−\frac{\mathrm{3}}{\mathrm{2}}<\mathrm{2}{k}<\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}}<{k}<\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow−\mathrm{0},…<{k}<\mathrm{0},…\:\Rightarrow{k}=\mathrm{0}\:\Rightarrow{x}=\frac{\pi}{\mathrm{2}} \\ $$$${case}\:\mathrm{2}\:\:−\pi<\frac{\pi}{\mathrm{6}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}<\pi\:\Rightarrow−\mathrm{1}<\frac{\mathrm{1}}{\mathrm{6}}\:+\:\frac{\mathrm{2}{k}}{\mathrm{3}}<\mathrm{1}\:\Rightarrow \\ $$$$−\frac{\mathrm{7}}{\mathrm{6}}<\frac{\mathrm{2}{k}}{\mathrm{3}}<\frac{\mathrm{5}}{\mathrm{6}}\:\Rightarrow−\frac{\mathrm{7}}{\mathrm{2}}<\mathrm{2}{k}<\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow−\frac{\mathrm{7}}{\mathrm{4}}<{k}<\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow−\mathrm{1},…<{k}<\mathrm{1},..\Rightarrow \\ $$$${k}\:\in\left\{−\mathrm{1},\mathrm{0},\mathrm{1}\right\}\: \\ $$$${k}=−\mathrm{1}\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{6}}−\frac{\mathrm{2}\pi}{\mathrm{3}}\:=\frac{−\mathrm{3}\pi}{\mathrm{6}}\:=−\frac{\pi}{\mathrm{2}} \\ $$$${k}=\mathrm{0}\:\Rightarrow{x}=\frac{\pi}{\mathrm{6}} \\ $$$${k}=\mathrm{1}\:\Rightarrow{x}=\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\:=\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

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