# cos^2(x) = 1/2 ... [o,2π)

• Nov 22nd 2010, 07:37 PM
mathnoob2
cos^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 PM
dwsmith
$\displaystyle \sqrt{cos^2x}=cos(x)=\sqrt{\frac{1}{2}}\rightarrow cos(x)=\frac{1}{\sqrt{2}}\rightarrow cos(x)=\frac{\sqrt{2}}{2}\rightarrow x=\frac{\pi}{4},\frac{7\pi}{4}$
• Nov 22nd 2010, 07:49 PM
pickslides
Quote:

Originally Posted by dwsmith
$\displaystyle cos(x)=\frac{\sqrt{2}}{2}=\frac{\pi}{4}$

I think you mean $\displaystyle \cos(x)=\frac{\sqrt{2}}{2}\implies x = \frac{\pi}{4}$

What about other solutions on $\displaystyle [0,2\pi)$ ??
• Nov 22nd 2010, 10:59 PM
mr fantastic
Quote:

Originally Posted by dwsmith
$\displaystyle \sqrt{cos^2x}=cos(x)=\sqrt{\frac{1}{2}}\rightarrow cos(x)=\frac{1}{\sqrt{2}}\rightarrow cos(x)=\frac{\sqrt{2}}{2}\rightarrow x=\frac{\pi}{4},\frac{7\pi}{4}$

The other case to consider is $\displaystyle \cos(x)=-\sqrt{\frac{1}{2}}$ and this is left for the OP to do.
• Nov 24th 2010, 10:16 AM
dhiab
• Nov 24th 2010, 09:23 PM
dwsmith
$\displaystyle \displaystyle \left[cos\left(\frac{3\pi}{4}\right)\right]^2=\left(\frac{-\sqrt{2}}{2}\right)^2=\frac{1}{2}$

Therefore, the angles that incorporate the negative values satisfy the equation.
• Nov 25th 2010, 01:37 AM
HallsofIvy
Quote:

Originally Posted by mathnoob2
Anyone have any idea how to solve it? Sqrt of both sides? But it's not a memory value angle...

As others have pointed out, yes, it is!

Quote:

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cos^2(x) = 1/2