Question

Hey, I'm reviewing for my midterm, and I'm wondering how I can solve 2 cos^2 2x = 1 over all real numbers and interval (0, 2pi]?

Answer 1

\[2 \cos^2(2x) =1 => \cos^2(2x)=\frac{1}{2}\]
\[=> \cos(2x)=\pm \frac{1}{\sqrt{2}} => \cos(2x)=\pm \frac{\sqrt{2}}{2}\]

Answer 2

now do you know when\[\cos(u)=\frac{\sqrt{2}}{2} \text{ or } \cos(u)=-\frac{\sqrt{2}}{2}\]

Answer 3

Well it could be both, because its in the interval (0, 2pi]?

Answer 4

Oh wait I think I see what you're trying to ask.

Answer 5

\[\cos(u)=\frac{\sqrt{2}}{2} \text{ when } u=\frac{\pi}{4}, \frac{5\pi}{4}\]
\[\cos(u)=\frac{-\sqrt{2}}{2} \text{ when } u=\frac{3\pi}{4},\frac{7\pi}{4}\]

Answer 6

Oh I see, thanks!

Answer 7

so since u=2x, then you can solve both of those equations I just wrote

Answer 8

for x

Answer 9

So the denominators become 8?

Answer 10

yep

Answer 11

so you understood everything?

Answer 12

yes! thank you <3

Answer 13

:)