find all solutions

• Apr 30th 2009, 09:32 AM
caseygaspar
find all solutions
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)
• Apr 30th 2009, 10:35 AM
Mush
Quote:

Originally Posted by caseygaspar
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)

$\displaystyle \sin(K) = \sin(P) \rightarrow K = \arcsin(\sin(P))$

And $\displaystyle \arcsin(\sin(P)) = P$
• Apr 30th 2009, 12:07 PM
sinewave85
Quote:

Originally Posted by caseygaspar
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)

To find solutions using graphs, you simply graph each side of the equation as a function: y = sin(x) and y=sin(1/2x). The solutions are the x-values at each point where the two graphs intersect.
• Apr 30th 2009, 01:14 PM
pflo
In these types of problems, you'll want all of your trig functions to be evaluated at the same number (in this case you've got $\displaystyle x$ AND $\displaystyle \frac{x}{2}$).

Use the half-angle identity for sine: $\displaystyle \sin(\frac{x}{2})={\pm}\sqrt{\frac{1-\cos(x)}{2}}$

So this root equals $\displaystyle \sin(x)$

Square both sides and you'll have $\displaystyle \frac{1-\cos(x)}{2}=\sin(x)^2$

Then use the pythagorean identity: $\displaystyle \sin(x)^2=1-\cos(x)^2$

You now have $\displaystyle \frac{1-\cos(x)}{2}=1-\cos(x)^2$

Put this equation into standard form for a quadratic equation and let $\displaystyle u=\cos(x)$

You now have $\displaystyle 0=2u^2-u-1$

Factor, solve, and put cosine back in for $\displaystyle u$ to get: $\displaystyle \cos(x)=-\frac{1}{2}$ and $\displaystyle \cos(x)=1$

Thus, all solutions are $\displaystyle x=\arccos(-\frac{1}{2})$ and $\displaystyle x=\arccos(1)$

Don't forget that the range of the arccosine function is limited, so there are three solutions to the original question in every period of the cosine.
• Apr 30th 2009, 01:19 PM
pflo
Quote:

Originally Posted by sinewave85
To find solutions using graphs, you simply graph each side of the equation as a function: y = sin(x) and y=sin(1/2x). The solutions are the x-values at each point where the two graphs intersect.

You could also get the equation equal to zero: $\displaystyle \sin(x)-\sin(\frac{x}{2})=0$ and graph this as a function. Then the zeros of the function are the solutions (the x-intercepts of $\displaystyle y=\sin(x)-\sin(\frac{x}{2})$).
• May 1st 2009, 06:43 AM
The Second Solution
Quote:

Originally Posted by Mush
$\displaystyle \sin(K) = \sin(P) \rightarrow K = \arcsin(\sin(P))$

And $\displaystyle \arcsin(\sin(P)) = P$

Actually it's $\displaystyle \sin(K) = \sin(P) \Rightarrow K = \arcsin(\sin(P)) + 2 n \pi$ or $\displaystyle K = [\pi - \arcsin(\sin(P))] + 2 n \pi$.