Answer
$0, \pi, 2 \pi$
Work Step by Step
We need to find the range for $x$ for the equation
$sinx=tanx$
$sinx-tanx=0$
$sinx-\frac{sinx}{cosx}=0$
$sinx(1-\frac{1}{cosx})=0$
Here, $sinx=0$ gives $x =0, \pi, 2\pi$
and $(1-\frac{1}{cosx})=0$ gives $x =0, 2\pi$.
Hence, $0, \pi, 2 \pi$