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Lösung 4.4:2c

Aus Online Mathematik Brückenkurs 1

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We get the full solution when we add multiples of <math>2\pi</math>,
We get the full solution when we add multiples of <math>2\pi</math>,
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{{Displayed math||<math>x = 0+2n\pi\qquad\text{and}\qquad x = \pi + 2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x = 0+2n\pi\qquad\text{and}\qquad x = \pi + 2n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.
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Note: Because the difference between <math>0</math> and <math>\pi</math> is a half turn, the solutions are repeated every half turn and they can be summarized in one expression,
Note: Because the difference between <math>0</math> and <math>\pi</math> is a half turn, the solutions are repeated every half turn and they can be summarized in one expression,
-
{{Displayed math||<math>x=0+n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x=0+n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.

Version vom 08:58, 22. Okt. 2008

There are two angles in the unit circle, x=0 and x=, whose sine has a value of zero.

We get the full solution when we add multiples of 2,

x=0+2nandx=+2n

where n is an arbitrary integer.


Note: Because the difference between 0 and is a half turn, the solutions are repeated every half turn and they can be summarized in one expression,

x=0+n

where n is an arbitrary integer.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:2c“