sin(theta/2)-cos(theta)=0

sin(theta/2)-cos(theta)=0     add cos(theta) to both sides and square both sides

[1 - cos(Θ]/2  =  cos²Θ    mutiply both sides by 2

1 - cos(Θ) =  2cosΘ        rearranging, we have

2cos²Θ + cos(Θ) - 1  =  0    factor

(2cos(Θ) - 1) (cos(Θ) +1 ) = 0

So. setting both factors to 0, we have

(2cos(Θ) - 1) = 0       add 1 to both sides

2cos(Θ) = 1               divide both sides by 2

cos(Θ) = 1/2        and this happens at  60 degrees  and at 300 degrees   on [0, 360]

And setiing the other factor to 0, we have

cos(Θ) +1  = 0   subtract  1 from both sides

cos(Θ)  = -1        and this happens at 180 degrees on [0, 360]

However, confining ourselves to [0, 360]......180 degrees is not a solution to the original equation .... (because we squared both sides, we sometimes get extraneous solutions in the interval !!)......however..... .-180 degrees would work ...!!!

Here's a graph of the solutions on [0, 360].....https://www.desmos.com/calculator/npmybx6bgv

sin(θ/2) - cos(θ/2)  =  0

--->   sin(θ/2)  =  cos(θ/2) 

--->   sin(θ/2) / cos(θ/2)  =  cos(θ/2) / cos(θ/2)

--->   tan(θ/2)  =  1

--->   θ/2  =  45° or   θ/2  =  135°

--->   θ  =  90°  or   θ  =  270°

Ok you had me confused - Gino answered the wrong question.  Easy done :)))