Question

How do you find all solutions of `2cos^2(θ)-1=0` to be `pi/4 +kpi` and `3pi/4 + kpi`? I keep getting `2kpi` instead of `kpi`

2`cos^2`θ-1=0
2`cos^2`θ=1
`cos^2`θ=`1/2`
cosθ=1/`sqrt2`=`sqrt2`/2
θ=±`pi`/4 +2k`pi`

θ=`pi`/4 +2`pi`k
or
θ=3`pi`/4 +2`pi`k

but my book says the answer is:

θ=`pi`/4 +k`pi`
or
θ=3`pi`/4 +k`pi`

Answer 1

`2cos^2(theta)-1 = 0`

`=>`cos^2(theta) = 1/2#

`=> cos(theta) = +-sqrt(1/2) = +-sqrt(2)/2`

Checking our unit circle, we find that `cos(theta) = sqrt(2)/2` at `theta = pi/4+2pik` and `theta=(7pi)/4 + 2pik`. For the negative value, we get `cos(theta)=-sqrt(2)/2` for `theta=(3pi)/4+2pik` and `theta=(5pi)/4+2pik`.

The problem with your given answer is that it does not account for solutions at which the angle is equivalent to `(5pi)/4` or `(7pi)/4`. Now let's see how the book was able to put it in the given form.

Noting that `pi/4` and `(5pi)/4` are offset by exactly `pi`, we know `(5pi)/4+2pik` will just be `pi/4+2pik+pi = pi/4+pi(2k+1)`. Thus, we can combine those answers to be `pi/4+pik`, as adding `pi` an even number of times will result in us landing on `pi/4` and adding `pi` an odd number of times will result in us landing on `(5pi)/4`.

Similarly, we can combine `(3pi)/4+2pik` and `(7pi)/4+2pik` into a single answer of `(3pi)/4+pik`, as an even number of rotations of `pi` will put us back on `(3pi)/4` and an odd number will put us on `(7pi)/4`.

The book stops there, but if we wanted to put the answer in its most concise form, we could just note that every answer is an integer number of rotations of `pi/2` from `pi/4`, and that any such rotation will land on a solution. Thus, the solution set could also be written as

`x=pi/4+(pi/2)k`