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**a4swe** $\displaystyle \cos(4v)=-\frac{1}{\sqrt{2}} \Leftrightarrow$

$\displaystyle -\cos(4v)=\frac{1}{\sqrt{2}} \Leftrightarrow$

$\displaystyle cos(4v+\pi)=\frac{1}{\sqrt{2}} \Leftrightarrow$

$\displaystyle 4v+\pi=+-\frac{\pi}{4}+n2\pi$

I:

$\displaystyle 4v+\pi=\frac{\pi}{4}+n2\pi \Leftrightarrow$

$\displaystyle v=-\frac{3\pi}{16}+\frac{n\pi}{2}$

II:

$\displaystyle 4v+\pi=-\frac{\pi}{4}+n2\pi \Leftrightarrow$

$\displaystyle v=-\frac{5\pi}{16}+\frac{n\pi}{2}$

Where I and II are my two solutions.

The answer given in my book are:

$\displaystyle +-\frac{3\pi}{16}+\frac{n\pi}{2}$

In this post I use a plus and a minus following eachother (like this: +-) to indicate that the following term can be either negative or positive.