# How to solve $\sqrt 2 \cos 2x = -1$?

I have this equation, $\sqrt{2} \cos 2x = -1$. I need all solutions between $[0,2\pi]$. I simplified that to $\cos 2x = -\frac{1}{\sqrt{2}}$.

I could just use a calculator and do $\arccos{-\frac{1}{\sqrt{2}}}$, to get angles for $2x$, but I need it in terms of $\pi$, and I also feel like I'm missing something simple that will allow me to find the solutions without one.

Should I just do it with a calculator? Or am I missing something here?

Thanks!

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You should work on knowing well that $\cos(\pi/4) = \sin(\pi/4) = \frac 1{\sqrt 2}$, and thus $\cos(5\pi/4) = -\frac 1{\sqrt 2}$. This will go a long way to knowing $\arccos \left(\frac 1{\sqrt 2}\right)$, etc. –  amWhy Jan 19 '14 at 18:43
Thats what I was missing...I learned $cos\frac{\pi}{4}$ as $\frac{\sqrt 2}{2}$. That's also true; somehow I didn't make the connection between those two values. Thanks! –  evamvid Jan 19 '14 at 18:53
You're welcome! –  amWhy Jan 19 '14 at 18:55

$$\cos (2x) = -\frac 1{\sqrt 2} \iff 2x = \arccos \left(-\frac{1}{\sqrt 2}\right)\iff x = \frac 12 \arccos \left(-\frac{1}{\sqrt 2}\right)$$
You'll find, then, that $$x=3\pi/8, \;5\pi/8,\; 11\pi/8,\; 13\pi/8$$
I get the procedure; thanks for that. What I'm still confused about is how you go from the last step to the answers shown. How do you get them in terms of $\pi$? To clarify, if you plug it into a calculator, it gives the answer in straight radians -- not in terms of $\pi$radians. –  evamvid Jan 19 '14 at 18:41
See my comment below your question. You'll want to memorize the foundational values of cosine and sine of of angles like $\pi/4, \pi/3, \pi/2, \pi, 3\pi/2, 0 = 2\pi$ –  amWhy Jan 19 '14 at 18:47
Letting $t=2x,$ you'll have $$\cos t=-1/\sqrt 2.$$ Since $0\le x\le 2\pi\iff 0\le t\le 4\pi,$ $$t=2x=3\pi/4, 5\pi/4, 11\pi/4, 13\pi/4.$$ Hence, $$x=3\pi/8, 5\pi/8, 11\pi/8, 13\pi/8.$$