## Calculus 8th Edition

$0, \pi, 2 \pi$
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 $cosx=1$ $x =0, 2\pi$ Hence, $0, \pi, 2 \pi$