cos^2(x) = 1/2 ... [o,2π)

**mathnoob2** · November 22nd 2010, 07:37 PM

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

**dwsmith** · November 22nd 2010, 07:40 PM

$\sqrt{cos^2x}=cos(x)=\sqrt{\frac{1}{2}}\rightarrow cos(x)=\frac{1}{\sqrt{2}}\rightarrow cos(x)=\frac{\sqrt{2}}{2}\rightarrow x=\frac{\pi}{4},\frac{7\pi}{4}$

**pickslides** · November 22nd 2010, 07:49 PM

Originally Posted by dwsmith

$cos(x)=\frac{\sqrt{2}}{2}=\frac{\pi}{4}$

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

What about other solutions on $[0,2\pi)$ ??

**mr fantastic** · November 22nd 2010, 10:59 PM

Originally Posted by dwsmith

$\sqrt{cos^2x}=cos(x)=\sqrt{\frac{1}{2}}\rightarrow cos(x)=\frac{1}{\sqrt{2}}\rightarrow cos(x)=\frac{\sqrt{2}}{2}\rightarrow x=\frac{\pi}{4},\frac{7\pi}{4}$

The other case to consider is $\cos(x)=-\sqrt{\frac{1}{2}}$ and this is left for the OP to do.

**dhiab** · November 24th 2010, 10:16 AM

Hello : $\sqrt{A^{2}}=\left | A \right |$

**dwsmith** · November 24th 2010, 09:23 PM

$\displaystyle \left[cos\left(\frac{3\pi}{4}\right)\right]^2=\left(\frac{-\sqrt{2}}{2}\right)^2=\frac{1}{2}$

Therefore, the angles that incorporate the negative values satisfy the equation.

**HallsofIvy** · November 25th 2010, 01:37 AM

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!

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cos^2(x) = 1/2

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