# How do you solve the given equation in the internal [0,2pi) for Tanx/2-sinx=0?

To solve this equation, use the half-angle formula for the tangent function, which greatly simplifies the expression.

Hello!

The equation is `tan ( x / 2 ) - sin x = 0 ` on `x in [ 0 , 2 pi ) . ` This interval is essential, because both functions `tan ( x / 2 ) ` and `sin x ` are `2 pi `-periodic.

But, while `sin x ` is defined for all values of `x , ` `tan ( x / 2 ) ` is not defined at `x / 2 = pi / 2 + n pi , ` where `n ` is an integer. So we note that `x != pi ` (the only such value at the given interval).

Use the half-angle formula for the tangent, `tan ( x / 2 ) = ( sin x ) / ( 1 + cos x ) . ` Then, the equation becomes `( sin x ) / ( 1 + cos x ) - sin x = 0 , ` or `( sin x ) / ( 1 + cos x ) ( 1 + cos x - 1 ) = 0 , ` or `( sin x cos x ) / ( 1 + cos x ) = 0 .`

The denominator is zero only at `x = pi , ` but we excluded this point already. The numerator is zero where `sin x = 0 , ` that is, at `x = 0 ` and `x = pi, ` and where `cos x = 0 , ` that is, at `x = pi / 2 ` and `x = ( 3 pi ) / 2 .`

This way, the solutions are `x = 0 , ` `x = pi / 2 ` and `x = ( 3 pi ) / 2 .`