Question

Question #13e76

Answer 1

`{theta | theta = pi/2 + kpi, " " theta = pi + 2kpi} " "" "" "k in ZZ`

Explanation:

One of the properties of `0` is that:

`ab = 0 " "` implies that `" " a=0 or b=0`

So our solutions are when:

`costheta = 0 " " or " " costheta+1 = 0`

`costheta = 0 " " or " " costheta = -1`

For `costheta` to be `0`, `theta` must be perpendicular to the `x`-axis. In other words, it must be either `pi/2` or `(3pi)/2`.

For `costheta` to be `-1`, `theta` must be `pi` since that is the lowest peak of the cosine curve.

Therefore, our solutions between `0` and `2pi` are:

`theta in {pi/2, pi, (3pi)/2}`

To include all solutions, we need to remember that adding `2pi` doesn't change the angle at all. So our solution set is:

`{theta | theta = pi/2 + 2kpi, " "theta = pi + 2kpi, " " theta = (3pi)/2 + 2kpi}`

To further simplify the solution set, we can combine `pi/2 + 2kpi` and `(3pi)/2 + 2kpi` into `pi/2 + kpi`.

`{theta | theta = pi/2 + kpi, " " theta = pi + 2kpi} " "" "" "k in ZZ`

Final Answer

Answer 2

Set both factors equal to 0.

Answer:

`theta = pi/2+npi; n in ZZ" [1]"`
`theta = pi+n2pi; n in ZZ" [2]"`

Explanation:

Given: `(cos(theta))(cos(theta)+1)=0`

The equation is 0 if either factor is 0:

`cos(theta) = 0 and cos(theta)+1 = 0`

`cos(theta) = 0 and cos(theta) = -1`

The cosine function is 0 at `theta = pi/2` and repeats at any integer multiple of `pi`:

`theta = pi/2+npi; n in ZZ" [1]"`

The cosine function is -1 at `theta = pi` and repeat at any integer multiple of `2pi`:

`theta = pi+n2pi; n in ZZ" [2]"`

Equations [1] and [2] are the answers.