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

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

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

$$sin\left(2x\right)=cosx\iff sin\left(2x\right)=sin(\frac{\pi }{2}-x)\iff 2x=\frac{\pi }{2}-x+2k\pi or$$$$2x=\frac{\pi }{2}+x+2k\pi \left(kfromZ\right)\iff 3x=\frac{\pi }{2}+2k\pi orx=\frac{\pi }{2}+2k\pi \iff$$$$x=\frac{\pi }{6}+\frac{2k\pi }{3}orx=\frac{\pi }{2}+2k\pi (k\in Z)$$

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

$$case1-\pi <\frac{\pi }{2}+2k\pi <\pi \Rightarrow -1<\frac{1}{2}+2k<1\Rightarrow -\frac{3}{2}<2k<\frac{1}{2}\Rightarrow$$$$-\frac{3}{4}<k<\frac{1}{4}\Rightarrow -0,...<k<0,...\Rightarrow k=0\Rightarrow x=\frac{\pi }{2}$$$$case2-\pi <\frac{\pi }{6}+\frac{2k\pi }{3}<\pi \Rightarrow -1<\frac{1}{6}+\frac{2k}{3}<1\Rightarrow$$$$-\frac{7}{6}<\frac{2k}{3}<\frac{5}{6}\Rightarrow -\frac{7}{2}<2k<\frac{5}{2}\Rightarrow -\frac{7}{4}<k<\frac{5}{4}\Rightarrow -1,...<k<1,..\Rightarrow$$$$k\in \{-1,0,1\}$$$$k=-1\Rightarrow x=\frac{\pi }{6}-\frac{2\pi }{3}=\frac{-3\pi }{6}=-\frac{\pi }{2}$$$$k=0\Rightarrow x=\frac{\pi }{6}$$$$k=1\Rightarrow x=\frac{\pi }{6}+\frac{2\pi }{3}=\frac{5\pi }{6}$$

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