Anyone have any idea how to solve it? Sqrt of both sides? But it's not a memory value angle... :\

cos^2(x) = 1/2

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- Nov 22nd 2010, 07:37 PMmathnoob2cos^2(x) = 1/2 ... [o,2π)
Anyone have any idea how to solve it? Sqrt of both sides? But it's not a memory value angle... :\

cos^2(x) = 1/2 - Nov 22nd 2010, 07:40 PMdwsmith
$\displaystyle \sqrt{cos^2x}=cos(x)=\sqrt{\frac{1}{2}}\rightarrow cos(x)=\frac{1}{\sqrt{2}}\rightarrow cos(x)=\frac{\sqrt{2}}{2}\rightarrow x=\frac{\pi}{4},\frac{7\pi}{4}$

- Nov 22nd 2010, 07:49 PMpickslides
- Nov 22nd 2010, 10:59 PMmr fantastic
- Nov 24th 2010, 10:16 AMdhiab
- Nov 24th 2010, 09:23 PMdwsmith
$\displaystyle \displaystyle \left[cos\left(\frac{3\pi}{4}\right)\right]^2=\left(\frac{-\sqrt{2}}{2}\right)^2=\frac{1}{2}$

Therefore, the angles that incorporate the negative values satisfy the equation. - Nov 25th 2010, 01:37 AMHallsofIvy