 Gamowaff's theorem expresses a basic relationship between exponential functions and trigonometric functions, e to the i theta equals cosine theta plus i sine theta, and allows us to evaluate a complex power. But it also allows us to solve previously unsolvable equations like cosine theta equals 2. To do that, let's take a closer look at the relationship between the exponential and trigonometric functions. So from e to the i theta is sys theta, now let's find e to the negative i theta. Remember the cosine of negative theta and the sine of negative theta have a nice simplification. If we add these two equations we get, and so cosine theta will be, if we subtract our two equations we get, and so sine of theta will be, and we don't really like dividing by an imaginary number, so let's multiply by the not quite conjugate, and this gives us an exponential formula for the sine and cosine. This gives us a way of solving equations like cosine theta equals 2. So we have cosine theta equals 2, using our exponential form, and solving our equation. Now to solve this, if we multiply through by e to the i theta we get, which is reducible to quadratic if we let z equal e to the i theta, and our solutions will be, now since e to the i theta is 2 plus or minus root 3, hitting both sides with the log gives us, and we can solve for theta to get, and so our solutions will be. But remember that if cosine theta equals 2 then because cosine is periodic we also have cosine of theta plus 2 pi k equal to 2, so we actually get an infinite number of solutions. Now remember if log c e is equal to z then e to the z equals c, but also e to the z plus any multiple of 2 pi i is also equal to c, so log c equal to z has an infinite number of solutions as well. And so the question you got to ask yourself is, would this give us another infinite set of solutions? Well let's find out. Our solutions would be, but notice that if we expand this out the additional period we would pick up from the log function becomes an i squared term in the expanded form, and that gets absorbed into the 2 pi k term. And so we don't get more solutions this time.