find all solutions

**caseygaspar** · April 30th 2009, 09:32 AM

hey guys I am a little confused on where to start! please help?
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)

**Mush** · April 30th 2009, 10:35 AM

Originally Posted by caseygaspar

hey guys I am a little confused on where to start! please help?
so find all solutions to this equation by using graphs...
sinx=sin(1/2x)

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

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

**sinewave85** · April 30th 2009, 12:07 PM

Originally Posted by caseygaspar

hey guys I am a little confused on where to start! please help?
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.

**pflo** · April 30th 2009, 01:14 PM

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.

**pflo** · April 30th 2009, 01:19 PM

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 $y=\sin(x)-\sin(\frac{x}{2})$ ).

**The Second Solution** · May 1st 2009, 06:43 AM

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 $K = [\pi - \arcsin(\sin(P))] + 2 n \pi$ .

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