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opgave d

\sin ^2 (x) = \frac{1}{2}
\sin (x) =  - \sqrt {\frac{1}{2}} of \sin (x) = \sqrt {\frac{1}{2}}
\sin (x) =  - \frac{1}{2}\sqrt 2 of \sin (x) = \frac{1}{2}\sqrt 2
x = 1\frac{1}{4}\pi  + k \cdot 2\pi of x = 1\frac{3}{4}\pi  + k \cdot 2\pi of x = \frac{1}{4}\pi  + k \cdot 2\pi of x = \frac{3}{4}\pi  + k \cdot 2\pi
x = \frac{1}{4}\pi  + k \cdot \frac{1}{2}\pi

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