# Solve the equation on the interval `(0,2pi).` `tan (x/2)=sin (x/2)`

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### 1 Answer

`tan (x/2)=(sin(x/2))/(cos(x/2)),` which equals `sin(x/2)` if `sin(x/2)=cos(x/2)sin(x/2),` which we factor as `sin(x/2)(1-cos(x/2))=0,` so the solutions occur whenever `sin(x/2)=0` or `cos(x/2)=1.`

`sin(x/2)=0` if and only if `x/2=piz,` where `z` is an integer, so `x=2piz` is a solution for any `z`, but ` `none of these solutions is in the interval `(0,2pi).`

`cos(x/2)=1` if and only if `x/2=2piz,` where `z` is an integer, so `x=4piz` is a solution for any `z.` However, these solutions are a subset of the ones we just found, and none of them are in the interval `(0,2pi)` either.

**The equation has no solution on `(0,2pi).` **