Solve the given equations in the interval [0,2pi). a. sin x/2 + cos x-1 = 0 b. tan x/2 - sin x=0

(a) On the interval [0,2pi), the solutions are 0,pi/3,(5pi)/3.

(b) On the interval [0,2pi), the solutions are pi/2, (3pi)/2, and 0.

Expert Answers

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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`

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



`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:


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


Multiplying by the denominator yields


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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