Question

Question #265f1

Answer 1

`pi/3` or `{5pi}/3`

Explanation:

Use the double angle formula
`cos (2A) = 1-2sin^2 A`
to rewrite the equation in the form
`0 = sin(x/2)- cos(x) = sin(x/2)-1+2sin^2(x/2)`

Substituting `t = sin(x/2)` this equation becomes
`2t^2+t-1 = 0 implies (2t-1)(t+1) = 0`

Since `x in [0,2pi)`, we have `x/2 in [0,pi)`, so that `t = sin(x/2) >= 0` Thus
`2t-1 = 0 implies t=1/2 implies sin(x/2) = 1/2`
Thus, either `x/2 = pi/6` or `x/2 = {5pi}/6`

Answer 2

`x=+-pi` or `x=+-pi/2`

Explanation:

Starting with `sin(x/2)-cosx=0`, use the half angle formula for `sin(x/2)`:
`sin(x/2)=sqrt((1-cosx)/2)` and rewrite the original equation:

`sqrt((1-cosx)/2)-cosx=0`
Add `cosx` to both sides: #sqrt((1-cosx)/2)-cosx=0 =>

`sqrt((1-cosx)/2)=cosx`
Square both sides to obtain `(1-cosx)/2=cos^2x`
Multiply both sides by 2, then put everything on one side:
`1-cosx=2cos^2x`
`2cos^2x +cosx-1=0`
You've essentially got a simple quadratic equation to factor now. If `n=cosx`, you could rewrite it as `2cos^2x -cosx+1=0`
=> `2n^2-n-1`, which factors as `(2n-1)(n+1)`.

So `2cos^2x +cosx-1=0` => `(2cosx -1)(cosx+1)=0`

Like solving for the zeros in any quadratic equation, set each factor equal to zero to find the solutions:
`2cosx-1=0`
`2cosx=1`

We know from the unit circle that `cosx=1/2` at `x=+-pi/3`, or equivalently, `x=+-60^o`

The next factor:
`cosx+1=0`
`cosx=-1`

Likewise, the unit circle shows us that `cosx=-1` where `x=+-pi`, and that `+-pi=+-180^o`

So, `x=+-pi` or `x=+-pi/3`
Or in degrees: `x=+-180^o` or `x=60^o`