Lösung 4.4:2e

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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,
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{{Displayed math||<math>5x = \frac{\pi}{6}\qquad\text{and}\qquad 5x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\,\textrm{.}</math>}}
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{{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,
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{{Displayed math||<math>5x = \frac{\pi}{6} + 2n\pi\qquad\text{and}\qquad 5x = \frac{5\pi}{6} + 2n\pi\,,</math>}}
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{{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,
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{{Displayed math||<math>x = \frac{\pi}{30} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{\pi}{6} + \frac{2}{5}n\pi\,,</math>}}
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{{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 \displaystyle 0\le 5x\le 2\pi, and using the unit circle shows that there are two of these,

\displaystyle 5x = \frac{\pi}{6}\qquad\text{and}\qquad 5x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\,\textrm{.}

We obtain the remaining solutions by adding multiples of \displaystyle 2\pi to the two solutions above,

\displaystyle 5x = \frac{\pi}{6} + 2n\pi\qquad\text{and}\qquad 5x = \frac{5\pi}{6} + 2n\pi\,,

where n is an arbitrary integer, or if we divide by 5,

\displaystyle x = \frac{\pi}{30} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{\pi}{6} + \frac{2}{5}n\pi\,,

where n is an arbitrary integer.

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