The exact solutions which lie in [0, 2π).
[tex]0,\:\pi ,\:\frac{\pi }{3},\:\frac{5\pi }{3}[/tex]
Given :
[tex]sin(2x)=sin(x)[/tex]
the exact solutions which lie in [0, 2π)
Subtract sin(x) from both sides
[tex]\sin \left(2x\right)-\sin \left(x\right)=0\\\\sin(2x)= 2 sinx cosx \\2sinxcosx-sinx=1\\factor \; out \;sin(x)\\\sin \left(x\right)\left(2\cos \left(x\right)-1\right)=0\\[/tex]
Set each factor =0 and solve for x
[tex]\sin \left(x\right)=0\quad \mathrm{or}\quad \:2\cos \left(x\right)-1=0\\sin(x)=0, x=0,\:x=\pi \\\\2cos(x)-1=0\\2cos(x)=1\\cos(x)=\frac{1}{2} \\x=\frac{\pi }{3},\:x=\frac{5\pi }{3}[/tex]
Exact solutions for the given equation is
[tex]0,\:\pi ,\:\frac{\pi }{3},\:\frac{5\pi }{3}[/tex]
Learn more : brainly.com/question/5093486
