Solve: $2x$ = $3\pi (1 -\cos x)$ I have really no idea how to do this . I have come to $x = 3\pi (\sin^2(x/2))$  . How can I solve this? In such type of problems how can I take help of the graph(wolfram alpha equation solver used graph)? Will I have to plot $2x$ & then the R.H.S ? Plz teach me the technique so that I can easily use graph in such problems.
 A: There are five roots:

and you can find good approximations by starting the Newton's method in a suitable neighbourhood of any root, but I doubt there is a closed-form expression for the second and fourth root. Anyway, by symmetry, we have:
$x_1=0,\quad x_3 = \frac{3\pi}{2},\quad x_5=3\pi.$
A: This are the kind of equation the solution of which cannot be expressed in terms of elementary functions.
But, basically what they ask you is to find the intersection of the line $y=2x$ with the curve $y=3\pi (1-\cos(x))$.
By inspection, you know that $x=0$ is one of the solution but you have other as shown in the graph Jack D'Aurizio posted.
If you want to know how Newton method would work for this problem, just post.
Added later on request
Many equations have solutions which cannot be expressed using elementary functions; as pointed by Jack D'Aurizio, this is the case for the solutions $x_2$ and $x_4$ of the equation you posted. In such cases, numerical methods should be used and one of the simplest root-finding method is Newton. I shall try to explain it to you using as simple words as possible.
So, we have to solve an equation $f(x)=0$. You know, that in the neighbourhood of a point $(x_0,y_0)$ which is on the curve, the tangent is very close to the curve which means that, very locally, the two functions ($f(x)$ for the curve and $g(x)$ for the tangent) are close to each other and that, if the point $(x_0,y_0)$ is sufficiently close to the root, the solutions of $f(x)=0$ and $g(x)=0$ should be close to eachother. So, using the definition of the tangent, function $g(x)$ simply write $$g(x)=f(x_0)+f'(x_0)(x-x_0)$$ and then the solution of $g(x)=0$ is simply given by $$x=x_0-\frac{f(x_0)}{f'(x_0)}$$ But this $x$ is not the solution of $f(x)=0$, so let us repeat the procedure until the predicted value of $x$ does not change anymore (for a given level of accuracy). This means that starting from a guess of the solution (say $x_0$), Newton method will permanently update it according to  $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
For a first and simple illustration, let us consider the case of $f(x)=x^2-3$ and we shall start iterating at $x_0=1$. We have $f'(x)=2x$ and so the iterative scheme will be $$x_{n+1}=x_n-\frac{x_n^2-3}{2x_n}=\frac{x_n^2+3}{2x_n}$$ and we start the process at $x_0=1$. So, we have $x_1=2$, $x_2=1.75$, $x_3=1.73214$, $x_4=1.73205$ which the solution for six significant figures.
Now, let us take the case of the second root of your equation $f(x)=2x-3\pi (1 -\cos x)$. By inspection, you could notice that there is a root which is cloase to $x_0=0.4$. So, using $f'(x)=2-3 \pi \sin(x)$, let us do the same as for the simple first case; now, Newton iterates are $x_1=0.433540$, $x_2=0.431060$, $x_3=0.431046$ which is the solution for six significant figures.
Now, let us take the case of the fourth root of your equation. By inspection, you could notice that there is a root which is cloase to $x_0=9$. So, let us do the same as for the simple first case; now, Newton iterates are $x_1=8.99364$, $x_2=8.99373$ which is the solution for six significant figures.
I tried to make these explanations as simple as possible; from a purely mathematical point of view, there are other serious concerns but I suppose that, at least for the presnt time, starting sufficiently close to the solution, you could apply this method with success (this is one of the reason for which a detailed look at the plot of the function is crucial). To give you and idea, suppose that, for the last example, we are lazy and start iterating at $x_0=10$ instead of $x_0=9$ as we did. In sucha case, the iterates will be $x_1=9.62578$, $x_2=9.47333$, $x_3=9.42930$, $x_4=9.42483$, $x_5=9.42478$ which is in fact $3 \pi$ that is to say the fifth solution; so, because of the bad choice of the starting point, we missed the fourth solution.
I hope and wish that these notes give you some ideas about Newton method (you could google for that and find good documentation in Wikipedia).
Let me know your feeling.
