# multiple solutions to equations involving sine

• Jan 31st 2008, 12:50 PM
pinion
multiple solutions to equations involving sine
hello
i'll get straight to the problem

i'm looking for non-numerical, non-iterative methods for finding the solutions to equations such as this:

sin(pi*2^x) = sin(pi*2^(1-x)) = 0, over the range [-1 < x < 2]

in words, i'm looking for where the given functions simultaneously intersect each other and the x-axis
so,
sin(pi * 2^x) = 0
sin(pi * 2^(1-x)) = 0

pi * 2^x = c1 * pi
pi * 2^(1-x) = c2 * pi
where c1 and c2 are integers >= 0

2^x = c1
2^(1-x) = c2

..ok so clearly i'm just transforming this a little to see if i jog someone's memory

i vaguely remember something in calculus that would lead to multiple solutions..however it somehow involved integrals..does anyone know what those types of calculus equations are called? how to solve them?

*edit*
oh can't calculus find the points of intersection of a parabola and a line? can i use a similar process on my equations?
• Jan 31st 2008, 05:02 PM
ecMathGeek
$sin(\pi2^x)=sin(\pi2^{1-x})=0$

As you pointed out, you are looking for instances where:
$2^x=c_1$
$2^{1-x}=c_2$
Where $c_1, c_2$ are natural numbers.

So, you need $x=log_2{c_1}$ AND $x=1-log_2{c_2}$ (for the same values of x), which implies that:
$log_2{c_1}=1-log_2{c_2}$
$c_1=2^{1-log_2{c_2}}$
$c_1=\frac{2}{c_2}$

Since $c_1$ and $c_2$ must be natural numbers but are related such that $c_1=\frac{2}{c_2}$, $c_2=2,1$ where $c_1=1,2$. In either case, $x=log_2{c_1}=0,1$ are the only solutions.
• Feb 1st 2008, 07:01 AM
pinion
Quote:

Originally Posted by ecMathGeek
Since $c_1$ and $c_2$ must be natural numbers but are related such that $c_1=\frac{2}{c_2}$, $c_2=2,1$ where $c_1=1,2$. In either case, $x=log_2{c_1}=0,1$ are the only solutions.

ok, but is there a way of finding $x=0,1$ without looking at a graph or checking $c_1=0$ $c_1=1$ $c_1=2$ $c_1=3$ ... ?

for instance, how is it you find the points where $y=3^x$ and $y=x+1.2$ intersect? i know i've done that before, so i think i can do something similar to find where two functions intersect each other and the x axis.

*edit*
well the intersection of a parabola and a line is fairly simple and straightforward..just use the quadratic formula. i'm positive i've had calculus problems which would lead to multiple solutions, however as i cannot remember their form i don't know if i could make use of that method.

now, in the case of finding the intersection of $y=3^x$ and $y=x+1.2$, if one could somehow rotate $y=3^x$ so that $y=x+1.2$ became the x-axis, then you would simply solve the rotated function for $y=0$, then un-rotate the solutions. in the case of my two functions involving sine it might be slightly more difficult to adjust one of them so that the other became the x-axis. perhaps a rather off-the-wall idea, but we'll see if it goes anywhere. any ideas?

i guess this doesn't have so much to do with calculus, i shall start a different thread.