The number of real solutions to $\displaystyle x^4 - x^3 = cosec^2(x) - cot^2(x)$ is?

-- ok, so i figured that cosec^2(x) = 1 + cot^2(x), so that brings it to

$\displaystyle x^4 - x^3 = 1$ if i am right?

so how many solutions are there then? 4?

EDIT: using a graphics calc, i graphed x^4 - x^3 -1 =0 and checked how many x intercepts there were, and so there were 2 real solutions.

But is there a way of doing this without the calculator?