How to solve $3x+\sin x = e^x$ How doese one solve this equation?
$$
3x+\sin x = e^x
$$
I tried graphing it and could only find approximate solutions, not the exact solutions.
My friends said to use Newton-Raphson, Lagrange interpolation, etc., but I don't know what these are as they are much beyond the high school syllabus. 
 A: A nice method to find an approximate solution is to successively cut intervals in half, as follows: let's first rewrite this as $$f(x) = 3x + \sin x - e^x = 0$$ Now pick two values, $a$ and $b$, such that $f(a) < 0$ and $f(b) > 0$. (You might have to make a few guesses before finding such values!) In this case, let's choose $a = 0$ and $b = 1$:
$$f(a) = 3(0)+\sin(0)-e^{0} = -1 < 0$$
$$f(b) = 3(1)+\sin(1)-e^{1} = 1.12... > 0$$
Now, because our function $f(x)$ is "smooth", there must be a solution somewhere between $a$ and $b$. Find the point halfway in between them, $\frac{0+1}{2} = 0.5$, and check to see whether it makes $f(x)$ positive or negative:
$$f(0.5) =3(0.5)+\sin(0.5)-e^{0.5} = 0.33... > 0$$
So, since $f(x)$ is positive here, set $0.5$ as the new value for $b$. Once again calculate the midpoint of $a$ and $b$ (in this case, $\frac{0+0.5}{2} = 0.25$), evaluate it in $f$, and so on. Continue until you have the precision you are looking for.
A: sorry
f(b)=3(1)+sin(1)−e1= is Not Equal to ---> 1.12...
f(b)=3(1)+sin(1)−e1= is Equal to  ----> 0.299170578
And
f(0.5)=3(0.5)+sin(0.5)−e0.5= is Not Equal to ---> 0.33...>0
f(0.5)=3(0.5)+sin(0.5)−e0.5= is Equal to  ---->   -0.1399947352
