Help with precal please!

Question

anonymous · Answer

okay?

anonymous · Answer

$$2\cos ^{2}x = 1$$

anonymous · Answer

yes it's solving trigonometric equations

anonymous · Answer

solve for cosx

anonymous · Answer

$$\left(\begin{matrix}2Cos ^{2}x \ 2\end{matrix}\right) = \frac{ 1 }{ 2 }$$

anonymous · Answer

$$Cos^2x = \frac{ 1 }{ 2 }$$

anonymous · Answer

$$x = \cos^{-1} (\frac{ 1 }{ 2 })$$

anonymous · Answer

Then using a calculator or something do the right side of the last step

anonymous · Answer

you're just supposed to simplyfy it down to cosx = whatever and then you go to the unit circle and use that to determine where that value occurs for cos and that's your answer

anonymous · Answer

oh shot I missed cos^2, ignore the last step

anonymous · Answer

Oh I see so in the end you have $$Cosx = \sqrt{\frac{ 1 }{ 2 }}$$

anonymous · Answer

I got it down to  cos ^2x = 1/2 but from there you would have to square both sides and square root of 1/2 isn't on the unit circle

anonymous · Answer

Any ideas?

anonymous · Answer

I'm looking through my notes, I just did this unit

anonymous · Answer

I'm pretty sure it's 60 degrees but I want to make sure

anonymous · Answer

Ok

anonymous · Answer

Actually I don't know.... I could try messaging someone to help you out.

anonymous · Answer

That would be great.  60 degrees is 1/2 but i need square root of 1/2 which isn't an option so I messed up somewhere

anonymous · Answer

yeah

anonymous · Answer

@TheSmartOne