Question

How do you find the general solutions for `cos (x/2)=sqrt2/2`?

Answer 1

`pi/2` and `(3pi)/2`

Explanation:

There are a few different methods to use, here.

We can see that the argument of our function is going to require the half angle formula. We can analyze it conceptually or mechanically.

Conceptually:

We know that for certain values of cos that our output will be `sqrt2/2`. If we wanted `cos(x)=sqrt2/2` then `x=(kpi)/4,` where K is an odd integer. From here, multiply your result by 2 (because it's an inverse operation to our argument) and you'll end up at `x=(kpi)/2`, where k is an odd integer. Bound the interval to `[0,2pi]` and we're left with `pi/2,(3pi)/2`

Mechanically, Isolate `cos(x)`:
`sqrt2/2=+-(sqrt(1-cos(x)/2))`

Square both side
`2/4=(1-cos(x)/2)-> 1/2=(1-cos(x)/2)`

Multiply both sides by 2
`1=1-cos(x)`

Subtract 1 from both sides to isolate `Cos(x)`
`0=cos(x)` (Because one side equals zero, we can drop the negative)

From here, we need to assess when `cos(x)=0`

Cos(x) is equal to zero at `pi/2, (3pi)/2` when bounded to `[0,2pi]`

Checking our work. `Cos((pi/2)/2)=cos(pi/4)=sqrt2/2`
and
`Cos(((3pi)/2)/2)=cos((3pi)/4)=sqrt2/2`

Answer 2

`x = pi/2 + 4kpi`
`x = (3pi)/2 + 4kpi`

Explanation:

`cos (x/2) = sqrt2/2`
Trig table and unit circle give 2 solutions for x/2 -->
1. `x/2 = pi/4 + 2kpi`
`x = pi/2 + 4kpi`
2. `x/2 = pi - pi/4 = (3pi)/4 + 2kpi`
`x = (3pi)/2 + 4kpi`