i keep on getting the wrong answer but cant work out where im going wrong

solve: between 0 and 360

3cos2 = 2cos^2

3(2cos^2 - 1) = 2cos^2

6cos^2 - 3 = 2cos^2

4cos^2 - 3 = 0

cos = sqr 3/4

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- Dec 8th 2008, 02:04 PMdnoCan someone please tell me where im going wrong
i keep on getting the wrong answer but cant work out where im going wrong

solve: between 0 and 360

3cos2 = 2cos^2

3(2cos^2 - 1) = 2cos^2

6cos^2 - 3 = 2cos^2

4cos^2 - 3 = 0

cos = sqr 3/4 - Dec 8th 2008, 03:42 PMmr fantastic
Where is the unknown? What you've posted is mathemetically meaningless.

I assume you mean $\displaystyle 3 \cos (2{\color{red}x}) = 2 \cos^2 {\color{red}x}$.

In which case your last line should be $\displaystyle \cos {\color{red}x} = {\color{red}\pm} \frac{\sqrt{3}}{2}$, that is, $\displaystyle \cos {\color{red}x} = \frac{\sqrt{3}}{2}$**and**$\displaystyle \cos x = {\color{red}-} \frac{\sqrt{3}}{2}$. - Dec 8th 2008, 04:56 PMdno
- Dec 8th 2008, 05:02 PMmr fantastic
- Dec 8th 2008, 05:22 PMdno