Question

What is the domain of `f(x) = sqrt(sinx)`?

Answer 1

`x in RR: [2npi, (2n+1)pi] forall n in ZZ`

Explanation:

`f(x) = sqrt(sinx)`

`f(x)` is real where `sinx>=0`

I.e. `[0, pi], [2pi, 3pi], .....` `-> +oo`
and `[-2pi, -pi]. [-4pi, -3pi]. ......` `-> -oo`

To generalise; `[2npi, (2n+1)pi] forall n in ZZ`

Hence the domain of `f(x)` is `x in RR: [2npi, (2n+1)pi] forall n in ZZ`

This can be more easily visualised from the graph of `f(x)` below.
graph{sqrt(sinx) [-14.24, 14.24, -7.12, 7.12]}