To solve this equation, I will use the identity $\tan x = \frac{\sin x}{\cos x}.$ Then I can rewrite the equation as $\frac{\sin x}{\cos x} = \sin x.$ Multiplying both sides by $\cos x,$ I get $\sin x = \sin x \cos x.$ Subtracting $\sin x \cos x$ from both sides, I get $\sin x - \sin x \cos x = 0.$ Factoring out $\sin x,$ I get $\sin x (1 - \cos x) = 0.$ Now I can use the zero product property and set each factor equal to zero. So $\sin x = 0$ or $1 - \cos x = 0.$ Solving for $x,$ I get $x = \sin^{-1}(0)$ or $x = \cos^{-1}(1).$ The inverse sine function gives me $x = 0, \pi, 2 \pi$ as solutions in the given interval. The inverse cosine function gives me $x = 0, 2 \pi$ as solutions in the given interval. But I don't want to repeat the same solutions twice, so I will only list each one once. Therefore, the final solutions are $x = 0, \pi, 2 \pi.$ 
Solve $\tan x = \sin x$ for $0 \le x \le 2 \pi.$  Enter all the solutions, separated by commas.