Lösung 4.4:6c

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If we use the trigonometric relation
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<center> [[Image:4_4_6c.gif]] </center>
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<math>\text{sin }\left( -x \right)=-\text{sin }x</math>, the equation can be rewritten as
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<math>\sin 2x=\sin \left( -x \right)</math>
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In exercise 4.4:5a, we saw that an equality of the type
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<math>\sin u=\sin v\quad </math>
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is satisfied if
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<math>u=v+2n\pi </math>
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or
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<math>u=\pi -v+2n\pi </math>
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where
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<math>n\text{ }</math>
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is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
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<math>2x=-x+2n\pi </math>
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or
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<math>2x=\pi -\left( -x \right)+2n\pi </math>
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i.e.
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<math>3x=2n\pi </math>
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or
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<math>x=\pi +2n\pi </math>
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The solutions to the equation are thus
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<math>\left\{ \begin{array}{*{35}l}
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x=\frac{2n\pi }{3} \\
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x=\pi +2n\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 11:54, 1. Okt. 2008

If we use the trigonometric relation \displaystyle \text{sin }\left( -x \right)=-\text{sin }x, the equation can be rewritten as


\displaystyle \sin 2x=\sin \left( -x \right)


In exercise 4.4:5a, we saw that an equality of the type


\displaystyle \sin u=\sin v\quad


is satisfied if


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


where \displaystyle n\text{ } is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy


\displaystyle 2x=-x+2n\pi or \displaystyle 2x=\pi -\left( -x \right)+2n\pi


i.e.


\displaystyle 3x=2n\pi or \displaystyle x=\pi +2n\pi


The solutions to the equation are thus


\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{2n\pi }{3} \\ x=\pi +2n\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_4.4:6c“