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$\cos2x=\frac1{\sqrt2}$ is the original problem, and I have to solve for $x$. However, I'm not sure what to do after I substitute the double angle formulas for $\cos2x$. I know that $\frac1{\sqrt2}$ can be rationalized to $\frac{\sqrt2}{2}$.

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The double-angle formula will make this more complicated. Since you know $\cos 2x$, you can figure out what $2x$ must be - it's an angle whose cosine is $\frac{\sqrt{2}}{2}$. Once you know $2x$, just divide by $2$ to find $x$.

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+1 for being the only person to write any English and to say that the double-angle formula isn't necessary. – symplectomorphic Jul 23 '14 at 12:55

Let $y = 2x$. We know that if $\cos y = \frac{\sqrt 2}{2}$, then $y = \pm \pi/4 + 2k\pi, \;k\in \mathbb Z$.

That means $$2x = \pm\frac{\pi}{4} + 2k\pi \iff x = \pm\frac{\pi}{8} + k\pi$$

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$$\cos 2x=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}$$ $$2x=\pm\frac{\pi}{4}+2k\pi$$ $$x=\pm\frac{\pi}{8}+k\pi,k\in\mathbb Z$$

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I guess mentioning what values k can assume will make it perfect answer. – Vikram Jul 23 '14 at 12:59

put $y=2x$, so: $\cos(y)=\frac{\sqrt{2}}{2}$, so, $y=\frac{\pi}{4}$ and $x=\frac{\pi}{8}$

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