Lösung 4.4:3b
Aus Online Mathematik Brückenkurs 1
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We obtain all solutions to the equation when we add integer multiples of <math>2\pi\, </math>, | We obtain all solutions to the equation when we add integer multiples of <math>2\pi\, </math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x = \frac{\pi}{5} + 2n\pi\qquad\text{and}\qquad x = \frac{4\pi}{5} + 2n\pi\,,</math>}} |
where ''n'' is an arbitrary integer. | where ''n'' is an arbitrary integer. | ||
Version vom 08:59, 22. Okt. 2008
We see directly that \displaystyle x = \pi/5 is a solution to the equation, and using the unit circle we can also draw the conclusion that \displaystyle x = \pi - \pi/5 = 4\pi/5 is the only other solution between \displaystyle 0 and \displaystyle 2\pi\,.
We obtain all solutions to the equation when we add integer multiples of \displaystyle 2\pi\, ,
| \displaystyle x = \frac{\pi}{5} + 2n\pi\qquad\text{and}\qquad x = \frac{4\pi}{5} + 2n\pi\,, |
where n is an arbitrary integer.
