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Question Number 132347 by Study last updated on 13/Feb/21

sin(2x+(π/4))−sin(2x+(π/3))=0 x=?

$${sin}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right)−{sin}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{0}\:\:\:{x}=? \\ $$

Answered by EDWIN88 last updated on 13/Feb/21

sin X−sin Y = 2cos (((X+Y)/2))sin (((X−Y)/2)) where { ((X=2x+(π/4))),((Y=2x+(π/3))) :} sin (2x+(π/4))−sin (2x+(π/3))=0 2cos (2x+((7π)/(24))).sin (−(π/(24)))=0 cos (2x+((7π)/(24)))=0 = cos (π/2) 2x = ± (π/2)−((7π)/(24))+2nπ x = ±(π/4)−((7π)/(48))+nπ ⇒x= { ((((5π)/(48))+nπ)),((−((19)/(48))+nπ)) :}

$$\mathrm{sin}\:\mathrm{X}−\mathrm{sin}\:\mathrm{Y}\:=\:\mathrm{2cos}\:\left(\frac{\mathrm{X}+\mathrm{Y}}{\mathrm{2}}\right)\mathrm{sin}\:\:\left(\frac{\mathrm{X}−\mathrm{Y}}{\mathrm{2}}\right) \\ $$$$\mathrm{where}\:\begin{cases}{\mathrm{X}=\mathrm{2x}+\frac{\pi}{\mathrm{4}}}\\{\mathrm{Y}=\mathrm{2x}+\frac{\pi}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{2cos}\:\left(\mathrm{2x}+\frac{\mathrm{7}\pi}{\mathrm{24}}\right).\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{24}}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\mathrm{2x}+\frac{\mathrm{7}\pi}{\mathrm{24}}\right)=\mathrm{0}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\mathrm{2x}\:=\:\pm\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{24}}+\mathrm{2n}\pi \\ $$$$\:\mathrm{x}\:=\:\pm\frac{\pi}{\mathrm{4}}−\frac{\mathrm{7}\pi}{\mathrm{48}}+\mathrm{n}\pi\:\Rightarrow\mathrm{x}=\begin{cases}{\frac{\mathrm{5}\pi}{\mathrm{48}}+\mathrm{n}\pi}\\{−\frac{\mathrm{19}}{\mathrm{48}}+\mathrm{n}\pi}\end{cases} \\ $$

Answered by Ar Brandon last updated on 13/Feb/21

sin(2x+(π/4))−sin(2x+(π/3))=0 ⇒sin(2x+(π/4))=sin(2x+(π/3)) { ((2x+(π/4)=2x+(π/3))),((2x+(π/4)=π−(2x+(π/3)))) :} ⇒4x=π−(π/4)−(π/3)=((5π)/(12)) [2π] ⇒ x=((5π)/(48)) [(π/2)]

$$\mathrm{sin}\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\left(\mathrm{2x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{0}\:\Rightarrow\mathrm{sin}\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{sin}\left(\mathrm{2x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\begin{cases}{\mathrm{2x}+\frac{\pi}{\mathrm{4}}=\mathrm{2x}+\frac{\pi}{\mathrm{3}}}\\{\mathrm{2x}+\frac{\pi}{\mathrm{4}}=\pi−\left(\mathrm{2x}+\frac{\pi}{\mathrm{3}}\right)}\end{cases}\:\Rightarrow\mathrm{4x}=\pi−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{3}}=\frac{\mathrm{5}\pi}{\mathrm{12}}\:\left[\mathrm{2}\pi\right] \\ $$$$\:\:\Rightarrow\:\:\mathrm{x}=\frac{\mathrm{5}\pi}{\mathrm{48}}\:\left[\frac{\pi}{\mathrm{2}}\right] \\ $$

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