1. ## find all solutions

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

2. Originally Posted by caseygaspar
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)
$\sin(K) = \sin(P) \rightarrow K = \arcsin(\sin(P))$

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

3. 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.

4. 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 $x$ AND $\frac{x}{2}$).

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

So this root equals $\sin(x)$

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

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

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

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

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

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

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

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.

5. 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: $\sin(x)-\sin(\frac{x}{2})=0$ and graph this as a function. Then the zeros of the function are the solutions (the x-intercepts of $y=\sin(x)-\sin(\frac{x}{2})$).

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

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