Question

How do you solve `sin(x/2) + cosx =0` in the interval [0, 2pi]?

Answer 1

We know that

`cosx=1-2sin^2(x/2)`

Hence

#sin(x/2)+cos(x)=0=>sin(x/2)+1-2sin^2(x/2)=0=> 2sin^2(x/2)-sin(x/2)-1=0#

Set `u=sin(x/2)` so we have

#2u^2-u-1=0=>u^2-u+u^2-1=0=>u(u-1)+(u-1)(u+1)=0=> (u-1)(2u+1)=0#

For `u=1` we get `sin(x/2)=1=>x/2=2kpi+pi/2=>x=4kpi+pi`

where `k` integer.

And for `u=-1/2` we get `sin(x/2)=-1/2=>x/2=2kpi-pi/6=>x=4kpi-pi/3 or x=4kpi+7pi/3`

But since our solutions belong to `[0,2pi]` the acceptable values are

`x=pi`