Answer:
[tex]x=0, \\x=\pi[/tex]
Step-by-step explanation:
Recall the trigonometric identity [tex]\sin 2x=2\sin x\cos x[/tex].
Therefore, given [tex]\sin 2x=2\sin x[/tex], rewrite the left side of the equation:
[tex]2\sin x\cos x=2\sin x[/tex]
Subtract [tex]\sin x[/tex] from both sides:
[tex]2\sin x\cos x-2\sin x=0[/tex]
Factor out [tex]2\sin x[/tex] from both terms on the left:
[tex]2\sin x(\cos x-1)=0[/tex]
We now have two cases:
[tex]\begin{cases}2\sin x=0, x=k\pi\text{ for }k\in \mathbb{Z}\\\cos x-1=0,x=k2\pi\text{ for }k\in \mathbb{Z}\end{cases}[/tex]
Since the problem stipulates that [tex]x[/tex] is in the interval [tex][0, 2\pi)[/tex], we have:
[tex]\text{For }x\in [0, 2\pi):\\\sin x=0\rightarrow \boxed{x=0, x=\pi}\\\cos x-1=0\rightarrow \boxed{x=0}[/tex]
Recall that square brackets mean inclusive and parentheses mean exclusive. Therefore, [tex]2\pi \notin [0, 2\pi)[/tex].
