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)
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.