Find the number of solutions of $x^4-x\sin(\pi x) - \cos^3(\pi x)=0$ We are required to find the number of solutions in $\mathbb{R}$ of the equation
$$x^4-x\sin(\pi x) - \cos^3(\pi x)=0 \tag{1}$$
This is a problem on the behaviour of functions and is likely based on some relevant properties of the derivatives of the above equation. In particular, finding the solutions explicitly is ruled out. 
I made the following observations:


*

*The left hand side of $(1)$ is even and since $x=0$ is not a solution only an even number of solutions exist. 

*The LHS involves an $x^4$ term, and so after a sufficiently large $x$, the $x^4$ term dominates and no solutions exist. It is not difficult to show that no solutions exist for $x>2$. 

 A: (Wanting to solve a post on integration that was deleted by the proposer, encounter this problem that has not received an answer yet. I intend to solve it and follow the first comment of Dr Sonnhard Grauner).
Notice that $f (x)= x^4-x\sin(\pi x) - \cos^3(\pi x)$ is even so we try with the values $x\ge 0$ only.
 Clearly $ x\sin(\pi x)\le x$; we find out $ x\sin(\pi x)= x$ for the values $x=\frac{1+4k}{2}\pi$ and note that for this values $\cos(\pi x)=0$ hence $f(x)\ge x^4-x\gt 1$ for $x\gt 1$;  furthermore $f(0)=-2$. Consequently $$f(x)=0\iff 0\lt x\lt 1\text{ for $x$ positive}$$
Taking the derivative $$f’(x)=4x^3-\sin(\pi x)-\pi x\cos(\pi x)+3\pi\cos^2(\pi x)\sin(\pi x)$$ we look at some approximation to a root of $f’(x)=0$. One has $$f’(0.38)\approx 0.12744775\gt 0\text{ and }f(0.38)\approx -0.38235035\\f’(0.40)\approx -0.18339251\lt 0\text{ and } f(0.40)\approx -038433110$$
Hence $$f’(x_1)=0 \text { for } 0.38\lt x_1\lt 0.40$$ Taking the value $x_1=0.39=\frac{0.38+0.40}{2}$ we get $$f’(0.39)\approx -0.03720276 \text{ and } f(0.39)\approx -0.38267702$$
Noting now that the function $f$ has only two oscillations for $x\in [0,1]$ we can deduce that the derivative in $x_1\approx 0.39$ approaches $0$ giving a local maximum near enough to $-0.38267702$ (i.e. distinct of $0$) and the other value for which the derivative is equal to $0$ corresponds to a minimum of $f(x)$ (so less than the precedent).
This ensures that there is a single real root $x_0\in [0,1]$.
We have $$f(0.7551)\approx -0.00055994\lt 0\\f(0.7553)\approx 0.00049442\gt 0$$ Thus we can take as a good approximation $$\color{red}{x_0\approx 0.7552}$$
Finally the only real roots are $\pm 0.7552$
