Can the equation sin(x) = 2 be solved?

When it comes to solving trigonometric equations, we need to always remember that functions like sin(x) or cos(x) are periodic and as such, there will be infinitely many values of x which satisfy the equation and therefore infinitely many solutions. For instance, the equation sin(x) = 0 does not only have the solution x = 0, but also the solutions x = π, x = 2π, x = 3π, x = -π, x = -2π, and so on, because sin(x) is zero at all of those points. Hence, we say that sin(x) = 0 has the solution x = nπ where n is some integer, either negative or positive.

The equation sin(x) = 2 seems to be solvable in a similar sense, but the issue is sin(x) is bounded by -1 and 1, meaning that no value of x will ever give you sin(x) = 2, or at least, no “real” value of x will satisfy the equation. But when one considers complex numbers, the possibilities suddenly open up. To see how this works, let us write sin(x) in exponential form using Euler’s formula (which I discussed in detail in my previous article: https://medium.com/@oscarnieves100/deriving-trigonometric-identities-using-complex-numbers-25ecd5392fb8):

from which we simply write:

and then, multiplying both sides by exp(ix) yields

or simply

If we look at this equation carefully, we notice that we can make the substitution