[Math] How to solve $ \sqrt 2 \cos 2x = -1$

algebra-precalculustrigonometry

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!

Best Answer

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, $$x=3\pi/8, 5\pi/8, 11\pi/8, 13\pi/8.$$

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