Lösung 4.4:5c

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For a fixed value of
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<center> [[Image:4_4_5c-1(2).gif]] </center>
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<math>u</math>, an equality of the form
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{{NAVCONTENT_START}}
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<center> [[Image:4_4_5c-2(2).gif]] </center>
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<math>\cos u=\cos v</math>
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{{NAVCONTENT_STOP}}
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is satisfied by two angles
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<math>v</math>
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in the unit circle:
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<math>v=u</math>
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and
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<math>v=-u</math>
[[Image:4_4_5_c.gif|center]]
[[Image:4_4_5_c.gif|center]]
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This means that all angles
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<math>v</math>
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which satisfy the equality are
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<math>v=u+2n\pi </math>
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and
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<math>v=-u+2n\pi </math>
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where
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<math>n\text{ }</math>
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is an arbitrary integer.
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Therefore, the equation
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<math>\cos 5x=\cos \left( x+{\pi }/{5}\; \right)</math>
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has the solutions
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<math>5x=x+\frac{\pi }{5}+2n\pi </math>
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or
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<math>5x=-x-\frac{\pi }{5}+2n\pi </math>
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If we collect
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<math>x\text{ }</math>
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onto one side, we end up with
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<math>\left\{ \begin{array}{*{35}l}
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x=\frac{\pi }{20}+\frac{1}{2}n\pi \\
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x=-\frac{\pi }{30}+\frac{1}{3}n\pi \\
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\end{array} \right.</math>
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(
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<math>n\text{ }</math>
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an arbitrary integer).

Version vom 11:18, 1. Okt. 2008

For a fixed value of \displaystyle u, an equality of the form


\displaystyle \cos u=\cos v


is satisfied by two angles \displaystyle v in the unit circle:


\displaystyle v=u and \displaystyle v=-u


This means that all angles \displaystyle v which satisfy the equality are


\displaystyle v=u+2n\pi and \displaystyle v=-u+2n\pi


where \displaystyle n\text{ } is an arbitrary integer.

Therefore, the equation


\displaystyle \cos 5x=\cos \left( x+{\pi }/{5}\; \right)


has the solutions


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

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

If we collect \displaystyle x\text{ } onto one side, we end up with


\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{20}+\frac{1}{2}n\pi \\ x=-\frac{\pi }{30}+\frac{1}{3}n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:5c“