Processing Math: Done
Lösung 4.4:2e
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
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This is almost the same equation as in exercise d. First, we determine the solutions to the equation when <math>0\le 5x\le 2\pi</math>, and using the unit circle shows that there are two of these, | This is almost the same equation as in exercise d. First, we determine the solutions to the equation when <math>0\le 5x\le 2\pi</math>, and using the unit circle shows that there are two of these, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>5x = \frac{\pi}{6}\qquad\text{and}\qquad 5x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\,\textrm{.}</math>}} |
[[Image:4_4_2_e.gif|center]] | [[Image:4_4_2_e.gif|center]] | ||
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We obtain the remaining solutions by adding multiples of <math>2\pi</math> to the two solutions above, | We obtain the remaining solutions by adding multiples of <math>2\pi</math> to the two solutions above, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>5x = \frac{\pi}{6} + 2n\pi\qquad\text{and}\qquad 5x = \frac{5\pi}{6} + 2n\pi\,,</math>}} |
where ''n'' is an arbitrary integer, or if we divide by 5, | where ''n'' is an arbitrary integer, or if we divide by 5, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x = \frac{\pi}{30} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{\pi}{6} + \frac{2}{5}n\pi\,,</math>}} |
where ''n'' is an arbitrary integer. | where ''n'' is an arbitrary integer. | ||
Version vom 08:58, 22. Okt. 2008
This is almost the same equation as in exercise d. First, we determine the solutions to the equation when 
5x2
| 6and5x=−6=65. |
We obtain the remaining solutions by adding multiples of
6+2nand5x=65+2n |
where n is an arbitrary integer, or if we divide by 5,
| 30+52nandx=6+52n |
where n is an arbitrary integer.
