Lösung 4.4:2e

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This is almost the same equation as in exercise d. First, we determine the solutions to the equation
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<center> [[Image:4_4_2e.gif]] </center>
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when
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<math>0\le \text{5}x\le \text{2}\pi </math>, and using the unit circle shows that there are two of these:
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<math>\text{5}x\text{ }=\frac{\pi }{6}</math>
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and
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<math>\text{5}x\text{ }=\pi -\frac{\pi }{6}=\frac{5\pi }{6}</math>.
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[[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
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<math>2\pi </math>
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to the two solutions above:
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<math>\text{5}x\text{ }=\frac{\pi }{6}+2n\pi </math>
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and
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<math>\text{5}x\text{ }=\frac{5\pi }{6}+2n\pi </math>
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(
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<math>n</math>
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an arbitrary integer),
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or if we divide by
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<math>\text{5}</math>:
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<math>x\text{ }=\frac{\pi }{30}+\frac{2}{5}n\pi </math>
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and
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<math>x\text{ }=\frac{\pi }{6}+\frac{2}{5}n\pi </math>
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(
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<math>n</math>
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an arbitrary integer).

Version vom 08:36, 1. Okt. 2008

This is almost the same equation as in exercise d. First, we determine the solutions to the equation when \displaystyle 0\le \text{5}x\le \text{2}\pi , and using the unit circle shows that there are two of these: \displaystyle \text{5}x\text{ }=\frac{\pi }{6} and \displaystyle \text{5}x\text{ }=\pi -\frac{\pi }{6}=\frac{5\pi }{6}.


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


\displaystyle \text{5}x\text{ }=\frac{\pi }{6}+2n\pi and \displaystyle \text{5}x\text{ }=\frac{5\pi }{6}+2n\pi ( \displaystyle n an arbitrary integer),

or if we divide by \displaystyle \text{5}:


\displaystyle x\text{ }=\frac{\pi }{30}+\frac{2}{5}n\pi and \displaystyle x\text{ }=\frac{\pi }{6}+\frac{2}{5}n\pi ( \displaystyle n an arbitrary integer).

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