How to solve this: two functions x = sin(x)/a - c

Question

I tried to solve this equation in different ways but no luck:
Find the number of points of two functions.
f(x) = sin(x)
y = a
at a given a. In this case lets say a = 0.15
sin(x) = ax
= 0.15x
x = sin(x)/0.15
???
Can any one help with this question?
this is the question in the exact words:
Write a C program that would read in the value a and write out all the solutions (i.e. roots) of the equation sin(x) = a*x
NOTE: This equation has for sure one solution: x = 0. However, for low values of a, the line representing the equation y = a*x is sufficiently close to horizontal line to cross the sine wave several times. Your program should calculate and print the number of roots and their values.
HINT: Suppose you are to solve an equation f(x) = 0, where f(x) is a continuous function. Suppose further that you can find two values xlow and xhigh such that f(xlow) < 0 and f(xhigh) > 0 and the function f(x) crosses the x-axis only once somewhere between these values. You could proceed using the so called "binary search" technique, as follows:
Estimate the solution x as x = (xlow + xhigh) / 2.0
If |f(x)| < epsilon then print x as solution and exit; else
If f(x) < 0 then substitute xlow = x else substitute xhigh = x
Go to (1)
Use epsilon = 0.001

Answer 1

You could proceed using the so called "binary search" technique
This is the key to the solution. Actually it is the solution. Let me draw two functions:
f(x) g(x)
/ --
/ --
/ --
/ --
/--
-*
--/
-- /
-- /
^ ^
| |
xlow xhigh
You have xlow and xhigh as an estimate of where f(x) crosses g(x). In your question, f(x) = ax and g(x) = sin(x).
First, let's see why xlow and xhigh make a good estimate. If you notice, at xlow, we have f(x) < g(x) and at xhigh we have f(x) > g(x). Since the functions are continuous, then there is some point somewhere in between where f(x) == g(x).
Now let's look at the middle-point between xlow and xhigh:
f(x) g(x)
/ --
/ --
/ --
/ --
/--
-*
--/
-- /
-- /
^ ^ ^
| | |
xlow xmid xhigh
Now at xmid, we have f(x) > g(x) (in this example). So:
f(xhigh) > g(xhigh)
f(xmid) > g(xmid)
f(xlow) < g(xlow)
Since between xmid and xlow, the functions change bigger-ness-ship (in other words, f(x) - g(x) changes its sign), then the answer is for sure between xlow and xmid (note that there could still be an even number of solutions between xmid and xhigh, but we can't really tell at this point).
So, if we assign xhigh = xmid, we would have:
f(x) g(x)
/--
-*
--/
-- /
-- /
^ ^
| |
xlow xhigh
But this is the same problem as before! Except that we shrank the possible location of the solution by half. Repeating we have:
f(x) g(x)
/--
-*
--/
-- /
-- /
^ ^ ^
| | |
xlow xmid xhigh
f(xhigh) < g(xhigh)
f(xmid) > g(xmid)
f(xlow) > g(xlow)
This time, the sign of f(x) - g(x) changes between xmid and xhigh, so we would do xlow = xmid to cut away the first half of the range which we are not interested in. We get:
f(x) g(x)
/--
-*
--/
^ ^
| |
xlow xhigh
Again, the same problem, except we shrank the range where the solution could possibly be by half.
Repeating this in a while loop, there would be some point where |f(xmid) - g(xmid)| becomes almost zero (say smaller than 0.000001 (or 1e-6) (also note that absolute value)). In that case, we stop searching and we say that that particular xmid is the answer. (See here for why we don't check for equality, but closeness).
There is still one problem. With your particular functions, there could be many cross-sections. How do we find xlow and xhigh? Well, we want the range [xlow, xhigh] to contain only one solution. So we can incrementally find those ranges and find the cross-section in between them.
Assuming a > 0 (and solutions for x > 0), this is how the graph would look like:
----- f(x) = ax
_ _ __*__ _ g(x) = sin(x)
/ \ /_*___----- / \ /
/ *____---*- \ / \ /
|---- | | | | | |
| | | | | |
\ / \ / \ /
\_/ \_/ \_/
So let's see where the solutions can be. For sure, it's not where sin(x) < 0. Then on [2kπ, 2kπ + π/2] and [2kπ + π/2, 2kπ + π] there could be one solution each. The initial [0, π / 2] may or may not have a solution depending on a. So the safest way is to enumerate all such ranges, calculate f(x) - g(x) for both xlow and xhigh and look at their signs. If the sign doesn't change, there is no solution and we can move on. If it does change, we perform the binary search above.
When does the algorithm end? We know that g(x) = sin(x) <= 1 and we know that for a > 0, f(x) = ax is always increasing. So when you have f(xlow) > 1, then for sure there are no more solutions.
The algorithm would thus be:
Main Algorithm:
for k = 0,1,...
xlow = 2kπ
xhigh = 2kπ + π/2
binary_search(xlow, xhigh)
xlow = 2kπ + π/2
xhigh = 2kπ + π
binary_search(xlow, xhigh)
binary_search:
if axlow-sin(xlow) and axhigh-sin(xhigh) have the same sign
return no result in this range
do
xmid = (xhigh + xlow) / 2
diff = axmid - sin(axmid)
if diff and axlow-sin(xlow) have the same sign
xlow = xmid
else
xhigh = xmid
while abs(diff) > epsilon
return xmid
For the cases where a < 0, the solution is similar (except the ranges change to the other half of the sin cycle). By the way, for each x that you find above, -x is also a solution!

Answer 2

Obviously, there can only be solutions for
-1 <= ax <= 1
The monotonicity of the difference function h(x)=sin(x)-ax is the same on all intervals between the roots of the derivative h'(x)=cos(x)-a, and they are also the local maxima and minima of this function. So use the points
arccos(a), 2*pi-arccos(a), 2*pi+arccos(a), 4*pi-arccos(a), 4*pi+arccos(a), ...
as long as they are smaller than 1/abs(a) to define the search intervals. Use inside each of them, as per the answer of Shahbaz, the bisection method, or better the Illinois variant of the regula falsi method.

Answer 3

A brutefore approach without any optimization (therefore a naive approach) would to run a for loop on x, x running from 0 to 2*PI, PI = 22.0/7.0, and loop increment by say 0.001, if you get abs(sin(x) - a*x) < epsilon, where epsilon say equals to some kow value like 0.001 (depends on your granularity), you get it! This would solve your equation sin(x) = a*x