Rewrite the equation as
[tex]\sin(x) = x^2-\pi^2 = (x-\pi)(x+\pi)[/tex]
If we choose [tex]x=\pm\pi[/tex], the right hand side becomes zero, and so does the left hand side.
If you graph the two functions [tex]f(x) = \sin(x)[/tex] and [tex]g(x) = x^2-\pi^2[/tex], you'll see that there are only two solutions, so the ones we found are the only ones.
