Postby owlpride » Sun Oct 02, 2011 12:48 am
There are 3 unique solutions (without counting multiplicities).
Notice that the left-hand side factors as x^2 * (x^2 - x sinx - cosx). The x^2 factor gives you a solution for x = 0. Let's focus on the other factor, f(x) = x^2 - x sinx - cosx. Notice that f is negative around a neighborhood of the origin and approaches + infinity as |x| gets large (since the x^2 term dominates). Hence f has at least two roots as its sign changes. Does it have any more?
Let's look at its critical points. f'(x) = 2x - x cosx = x * (2 - cosx) = 0 only at x = 0. (That's a local minimum of f because f' < 0 for x < 0 and f' > 0 for x > 0.) Since f has only a single critical point, it will have at most two roots (there's at least one critical point between any two roots). But we already found two roots. So we are done.