1. We are asked to solve `sin(x/2)+cos(x)-1=0` .

First, we use the half-angle formula for sine: `sin(x/2)=(-1)^(|_ (x/(2 pi)) _|)sqrt((1-cos x)/2)`.

So, `sqrt((1-cosx)/2)+cosx-1=0`
`sqrt((1-cosx)/2)=1-cosx`.

** The reason we can be so cavalier about the sign on the root is that we will square both sides of the equation.**

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

`1-cosx=2-4cosx+2cos^2x`

`2cos^2x-3cosx+1=0 => cosx=1/2` or `cosx=1`

`cosx=1/2 => x=pi/3,(5pi)/3` while `cosx=1 => x=0` on `[0,2pi)`

(See attachment to see that these are the solutions.)

2. Solve `tan(x/2)-sinx=0" on "[0,2pi)`.

As above, we could use the half-angle formula for tangent, but there is another way. First recognize that `tan theta=(sin theta)/(cos theta)`, so we can rewrite as:

`(sin(x/2))/(cos(x/2))-sinx=0`.

Now we can use the double-angle formula for sine to rewrite sin x as `2sin(x/2)cos(x/2)` so that

`(sin(x/2))/(cos(x/2))=2sin(x/2)cos(x/2)`.

Multiplying by the denominator yields

`sin(x/2)=2sin(x/2)cos^2(x/2)`.

On the interval `[0,2pi) sin(x/2)=0 => x=0`, and that is a solution. Now we can divide by `sin(x/2)` without losing a root and assuming `x ne 0` to get:

`cos^2(x/2)=1/2 => cos(x/2)=+- sqrt(2)/2`.

Then `x=pi/2, (3pi)/2`.

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