[tex]\sin2x-\sin x=2\sin x\cos x-\sin x=\sin x(2\cos x-1)=0[/tex]
[tex]\implies\sin x=0\text{ or }2\cos x-1=0[/tex]
From the first equation, we get two possible solutions at [tex]x=0[/tex] and [tex]x=\pi\approx3.14[/tex].
From the second equation, we get
[tex]2\cos x-1=0\implies\cos x=\dfrac12[/tex]
which has exact solutions at [tex]x=\dfrac\pi3[/tex] and [tex]x=\dfrac{5\pi}3[/tex]. Since [tex]\pi\approx3.14[/tex], we can approximate these solutions by [tex]x=\dfrac\pi3\approx1.05[/tex] and [tex]x=\dfrac{5\pi}3\approx5.24[/tex].
