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