Question

How do you solve the equation `cosZ=0.98`?

Answer 1

`Z = cos^(-1)(0.98) + 2pi*n`, `n in ZZ`.

or

`Z=- cos^(-1)(0.98) + 2pi*n`, `n in ZZ`.

Explanation:

Since `Z>0` we know that `Z` could come from QI or QIV. We'll find `hatZ`, the reference angle for `Z`:

`cos(Z)=0.98 rarr hatZ=cos^(-1)(0.98)`

For the QI angle, we have `Z = cos^(-1)(0.98) + 2pi*n`, `n in ZZ`.

For QIV, we have `Z = 2pi - cos^(-1)(0.98) + 2pi*n`, `n in ZZ`, which could be rewritten: `- cos^(-1)(0.98) + 2pi*n`, `n in ZZ`.