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

Solve it in ]−π;π] sin(2x)=cos(x)

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

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

sin(2x)=cosx ⇔ sin(2x)=sin((π/2)−x) ⇔2x=(π/2)−x+2kπ or 2x =(π/2)+x +2kπ ( kfrom Z) ⇔3x =(π/2) +2kπ or x=(π/2)+ 2kπ ⇔ x =(π/6) +((2kπ)/3) or x=(π/2) +2kπ (k∈ Z)

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

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

case 1 −π <(π/2)+2kπ <π ⇒−1<(1/2) +2k<1 ⇒−(3/2)<2k<(1/2) ⇒ −(3/4)<k<(1/4) ⇒−0,...<k<0,... ⇒k=0 ⇒x=(π/2) case 2 −π<(π/6) +((2kπ)/3)<π ⇒−1<(1/6) + ((2k)/3)<1 ⇒ −(7/6)<((2k)/3)<(5/6) ⇒−(7/2)<2k<(5/2) ⇒−(7/4)<k<(5/4) ⇒−1,...<k<1,..⇒ k ∈{−1,0,1} k=−1 ⇒x =(π/6)−((2π)/3) =((−3π)/6) =−(π/2) k=0 ⇒x=(π/6) k=1 ⇒x=(π/6)+((2π)/3) =((5π)/6)

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

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