The number of real roots of the equation $e^x+e^{-x}=2\sin(x^3)$ 
Find the number of real roots of the equation $e^x+e^{-x}=2\sin(x^3)$.

I got the answer as infinite number of roots. 
My attempt:
$e^{2x}+1=2e^x\sin(x^3)$
The range of the right hand side expression is approximately $[-5.4e^x,5.4e^x]$.
There are infinite values of $x$, which satisfy the above equation.
Am I correct? How can we find the roots of such types of equations?
 A: Since the $LHS$ is the sum of the form $y+\frac{1}{y}$, where $y$ is positive, hence the minimum value of $LHS$ is $2$ (By AM-GM inequality), which is attained when $y=1$. Also, the maximum value of $RHS$ is $2$.
Hence, solution is $x$ such that $e^x=1$ and $2sin(x^3)=2$. Since, the only solution of the former is 0 and which doesn't satisfy latter, hence no solution.
Hope it is helpful:)
A: Note that 


*

*We have $e^x + e^{-x}  \geq 2$ for all $x$ (why? Take $e^{-x} = y$ and rearrange $(y-1)^2 \geq 0$ to the given form)

*We have $2\sin (x^3) \leq 2$ for all $x$.(why?)
Thus if they are equal then we must have equality in both expressions above. This leads to $\sin(x^3) = 1$ and $y = e^x = 1$ , in other words $x=0$. 
One checks that this does not work, hence we have no solution.

How can we find the roots of such types of equations?  EDIT coming up.
EDIT : IN general, to show the existence of a real root, transfer all the terms to one side, and then you are looking at a situation where you are finding the roots of some function $f$, like I did above.
Now, I will try to give a good answer for the case of $f$ being continuous. We have one nice fact definitely, which we repeatedly use.

If $f(x) < 0$ for some $x$ and $f(y) > 0$ for some $y$, then $f$ has a root between $x$ and $y$. (Bolzano)

So one idea would be to look for numbers where you can be sure that $f$ is negative, and where you can be sure that $f$ is positive. Also, studying the behaviour of $f$ for very large and very small values of $x$ helps find roots since in the limit we may be able to guarantee that $f$ is positive/negative, so we can still apply IVT but not with a determined $x$ or $y$. 
Example : This is used to show that every polynomial of odd degree has a real root.

If $f$ is differentiable, then we can play around with the derivative. The most important result here would be Rolle's theorem, and the theory of maxima-minima.

Basically, Rolle's theorem tells you that between any two real roots of the function, there is a real root of the derivative. Now, if the derivative is nice, you can find all its real roots, say there are $n$ of them. Then, guaranteed, $f$ cannot have more than $n+1$ real roots.
For the maxima-minima, find the points of extremum by considering the roots of the derivative. These points can either be candidates for using Bolzano's theorem, or can be used to show that the graph lies above or below some vertical line for some period.

For example, here the derivative , which is $e^x - e^{-x} - 6x^2 \cos(x^3)$ is far too complicated to work with, so we prefer not to use it.
The method used then, is to show that $f$ breaks into two parts (your question even indicates that) that have incompatible ranges. I think the question was even phrased so that this could be spotted.
In general, however, most problems of this type (which are posed in exams/tests and not "unsolved/hugely difficult problems") can be solved using the techniques I have indicated. In fact, that's one of the reasons why calculus is so useful, in providing inequalities that cannot be proven with ordinary algebraic methods.
A: None.
The equation reads
$$\cosh x=\sin x^3.$$
But for $x\ne0$, $\cosh x>1.$
