How to solve for $x$ in $x(x^3+\sin x \cos x)-\sin^2 x =0$? 
How do I solve for $x$ in $$x\left(x^3+\sin(x)\cos(x)\right)-\big(\sin(x)\big)^2=0$$

I hate when I find something that looks simple, that I should know how to do, but it holds me up.  
I could come up with an approximate answer using Taylor's, but how do I solve this?  
(btw, WolframAlpha tells me the answer, but I want to know how it's solved.)

Using the identity $\cos x=1-2\sin^2(x/2)$ and introduccing the function ${\rm sinc}(x):={\sin x\over x}$ we can rewrite the given  function $f$  in the following way:
$$f(x)=x^2\left(x^2\left(1-{1\over2}{\rm sinc}(x){\rm sinc}^2(x/2)\right)+{\rm sinc}(x)\bigl(1-{\rm sinc}(x)\bigr)\right)\ .\qquad(*)$$
Now ${\rm sinc}(x)$ is $\geq0$ on $[0,\pi]$ and of absolute value $\leq1$ throughout. By distinguishing the cases $0<x\leq\pi$ and $x>\pi$ it can be verified by inspection that $f(x)>0$ for $x>0$. Since $f$ is even it follows that $x_0=0$ is the only real zero of $f$.
[One arrives at the representation $(*)$ by expanding the simple functions appearing in the given $f$ into a Taylor series around $0$ and grouping terms of the same order skillfully.]
