\begin{deloppgaver}
  \delo
    \begin{align*}
      cos(x) &= -1 \\
      x &= acos(-1) \\
      x &= \pm \pi + 2n\pi
    \end{align*}

    \begin{minipage}{0.42\textwidth}
      \begin{graphbox}
        \input{figures/4a.tex}
      \end{graphbox}
    \end{minipage}


    Ettersom forskjellen mellom $\pi$ og $-\pi$ er $2\pi$ som er et element i $2n\pi$, kan vi slå sammen svarene og si at

    \[ x = \pi + 2n\pi \]

  \delo
    \[cos(2x)=1-2sin^2(x)\]

    Vi substituerer $cos(2x)=cos^2(x)-sin^2(x)$ og $cos^2(x)+sin^2(x)=1$

    \begin{align*}
      cos^2(x)-sin^2(x) &= cos^2(x)+sin^2(x) - 2sin^2(x) \\
      cos^2(x) &= cos^2(x)+2sin^2(x) - 2sin^2(x) \\
      cos^2(x) &= cos^2(x) 
    \end{align*}
  
  \delo
    \[cos(\frac{\pi}{8})\]

    Vi bruker

    \[ sin(2a) = 2 sin(a)cos(a) \]

    Hvor $a = \frac{\pi}{8}$

    \begin{align*}
      sin\left(\frac{\pi}{4}\right) &= 2sin\left(\frac{\pi}{8}\right)cos\left(\frac{\pi}{8}\right) \\[1em]
      cos\left(\frac{\pi}{8}\right) &= \frac{sin\left(\frac{\pi}{4}\right)}{2sin\left(\frac{\pi}{8}\right)} \\[1em]
        &= \frac{\frac{1}{\sqrt{2}}}{2sin\left(\frac{\pi}{8}\right)}
    \end{align*}

\end{deloppgaver}