Lösung 4.4:3b

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We see directly that
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<center> [[Image:4_4_3b.gif]] </center>
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<math>x=\frac{\pi }{5}</math>
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is a solution to the equation, and using the unit circle we can also draw the conclusion that
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<math>x=\pi -\frac{\pi }{5}=\frac{4\pi }{5}</math>
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is the only other solution between
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<math>0</math>
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and
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<math>\text{2}\pi </math>.
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[[Image:4_4_3_b.gif|center]]
[[Image:4_4_3_b.gif|center]]
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We obtain all solutions to the equation when we add integer multiples of
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<math>\text{2}\pi </math>,
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<math>x=\frac{\pi }{5}+2n\pi </math>
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and
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<math>x=\frac{4\pi }{5}+2n\pi </math>
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where
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<math>n</math>
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is an arbitrary integer.

Version vom 09:43, 1. Okt. 2008

We see directly that \displaystyle x=\frac{\pi }{5} is a solution to the equation, and using the unit circle we can also draw the conclusion that \displaystyle x=\pi -\frac{\pi }{5}=\frac{4\pi }{5} is the only other solution between \displaystyle 0 and \displaystyle \text{2}\pi .


We obtain all solutions to the equation when we add integer multiples of \displaystyle \text{2}\pi ,


\displaystyle x=\frac{\pi }{5}+2n\pi and \displaystyle x=\frac{4\pi }{5}+2n\pi


where \displaystyle n is an arbitrary integer.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:3b“