The equation $\displaystyle 2-cos^2\theta=Acos2\theta$ where $\displaystyle A$ is a constant, has a root $\displaystyle \theta=30^o$. Find all the other roots such that $\displaystyle \theta$ is between $\displaystyle 0^o$ and $\displaystyle 360^o$

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- Sep 27th 2010, 03:23 AMPunchRoots on an equation
The equation $\displaystyle 2-cos^2\theta=Acos2\theta$ where $\displaystyle A$ is a constant, has a root $\displaystyle \theta=30^o$. Find all the other roots such that $\displaystyle \theta$ is between $\displaystyle 0^o$ and $\displaystyle 360^o$

- Sep 27th 2010, 05:10 AMmathaddict
- Sep 27th 2010, 06:26 AMPunch
Yes i did, A = 1.549

- Sep 28th 2010, 05:12 AMmathaddict
- Sep 30th 2010, 04:30 AMPunch
- Sep 30th 2010, 05:32 AMmathaddict
- Sep 30th 2010, 05:38 AMPunch
- Sep 30th 2010, 05:45 AMmathaddict
It seems like you are unsure how to solve s trigonometric equation. Review your class notes which should have covered that to get some basic understanding on this before you get back to this question.

Anyways, in case if it's the square of the cosine which is confusing you,

$\displaystyle \cos b =\pm \sqrt{\frac{A+2}{2A+1}}$