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)

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- Apr 30th 2009, 09:32 AMcaseygasparfind all solutions
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) - Apr 30th 2009, 10:35 AMMush
- Apr 30th 2009, 12:07 PMsinewave85
- Apr 30th 2009, 01:14 PMpflo
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 PMpflo
- May 1st 2009, 06:43 AMThe Second Solution