How can I solve $x\cos(x)=\pi$ without looking a graph $$x\cos(x)=\pi$$
Also, when looking at a graph, $\pi$ has the value $3.14$ right?
I can guess that if $x=-\pi$ then the equation has a solution. 
 A: In the first place, there is no hope for an analytical solution because $x$ appears both outside and inside of a trigonometric function. There is indeed the simple solution $x=-\pi$, but this is "accidental", you won't find a formula for the other roots.
The trick to address such difficult equations is to find the extrema of the function, because you are sure that there is at most one solution of $f(x)=y$ between successive minima and a maxima (unless there are discontinuities), and they must be of opposite sign.
To locate the extrema, we need to cancel the derivative,
$$f'(x)=\cos(x)-x\sin(x)=0.$$
This equation isn't much more appetizing than the first and we seem to be stuck. Anyway, as $x=0$ isn't a solution, we can solve
$$g(x)=\cot(x)-x=0$$ instead.
Now, taking the derivative,
$$g'(x)=-\frac1{\sin^2(x)}-1=0$$ has no solution as the LHS is strictly negative, and the function $g$ has no extrema.
Anyway, there are discontinuities as $g$ has vertical asymptotes for $x=k\pi$, and it is monotonic in all intervals $(k\pi,k\pi+\pi)$, running from $\infty$ to $-\infty$. So in every such interval, the function $g$ and the derivative $f'$ have exactly one root, and the initial function $f$ has exactly one extremum.
It just remains to check if there is a change of sign between 
$$f(k\pi)=k\pi\cos(k\pi)-\pi=((-1)^kk-1)\pi$$ and $$f(k\pi+\pi)=(k\pi+\pi)\cos(k\pi+\pi)-\pi=(-(-1)^{k}(k+1)-1)\pi.$$
The sequence is 
$$\cdots4,-5,2,-3,0,-1,-1,1,-2,3,-4,\cdots$$
(times $\pi$), and the sign changes for every $k$ but $k=-1$, corresponding to the exact root $x=-\pi$, and $k=0$, no root.
In conclusion, every interval $(k\pi,k\pi+\pi)$, for $k<-1$ and $k>0$ contains a single root (to be computed by numerical methods), and there is an extra root at $x=-\pi$.
(Also notice that for growing $k$, the equation $\cos(x)=\frac\pi x$ tends to $\cos(x)=0$, which you know how to solve :) A slightly better approximation is obtained by assuming the multiplicative $x$ to remain constant in the interval, $x\approx(k+\frac12)\pi$), and solve $\cos(x)=\frac1{k+\frac12}$.)
