Question

How do you solve `e^sinx=1` in the interval [0,360]?

Answer 1

`e^x` is an increasing function over allspace (`d/(dx)[e^x] = e^x`, which is always nonnegative), so it must be that `e^u` crosses `y = 1` once, i.e. when `u = 0`. To prove this:

`lne^(sinx) = ln1`

`sinx = 0`

And we would have that `sinx = 0` in `[0,2pi]` when `x = {0,pi,2pi}`. In general, because of the periodicity of `sin`, we would then have that

`color(blue)(x = npi)` where `n in ZZ` (where `n` is an integer).