Find an angle x in radians such that 8pi<x<9pi and cosx=1/2

What is this question asking for? I came up with pi/3, but I don't think that's in between 8 and 9 pi. Can anyone explain? Thanks

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- Feb 28th 2009, 06:38 AMcaptaintoast87finding and angle x in radians?
Find an angle x in radians such that 8pi<x<9pi and cosx=1/2

What is this question asking for? I came up with pi/3, but I don't think that's in between 8 and 9 pi. Can anyone explain? Thanks - Feb 28th 2009, 06:59 AMe^(i*pi)
Remember that $\displaystyle \cos{x}$ is periodic with a a period of $\displaystyle 2\pi$ which means you get your original answer of $\displaystyle \frac{\pi}{3}$ and add 4 lots of $\displaystyle 2\pi$ (this is 4 periods) to give you an answer of $\displaystyle 8\pi + \frac{\pi}{3} = \frac{25\pi}{3}$