Lösung 4.4:2a

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{{NAVCONTENT_START}}
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We draw a unit circle and mark on those angles on the circle which have a
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<center> [[Image:4_4_2a-1(2).gif]] </center>
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<math>y</math>
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{{NAVCONTENT_STOP}}
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-coordinate of
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{{NAVCONTENT_START}}
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<math>{\sqrt{3}}/{2}\;</math>, in order to see which solutions lie between
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<center> [[Image:4_4_2a-2(2).gif]] </center>
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<math>0</math>
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{{NAVCONTENT_STOP}}
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and
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<math>2\pi </math>.
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[[Image:4_4_2_a.gif|center]]
[[Image:4_4_2_a.gif|center]]
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In the first quadrant, we recognize
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<math>x={\pi }/{3}\;</math>
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as the angle which has a sine value of
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<math>{\sqrt{3}}/{2}\;</math>
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and then we have the reflectionally symmetric solution
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<math>x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}</math>
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in the second quadrant.
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Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of
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<math>2\pi </math>
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<math>x=\frac{\pi }{3}+2n\pi </math>
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and
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<math>x=\frac{2\pi }{3}+2n\pi </math>
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where
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<math>n</math>
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is an arbitrary integer.
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NOTE: when we write that the complete solution is given by
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<math>x=\frac{\pi }{3}+2n\pi </math>
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and
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<math>x=\frac{2\pi }{3}+2n\pi </math>,
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this means that for every integer
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<math>n</math>, we obtain a solution to the equation:
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<math>\begin{array}{*{35}l}
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n=0 & x=\frac{\pi }{3} & x=\frac{2\pi }{3} \\
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n=-1 & x=\frac{\pi }{3}+\left( -1 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -1 \right)\centerdot 2\pi \\
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n=1 & x=\frac{\pi }{3}+1\centerdot 2\pi & x=\frac{2\pi }{3}+1\centerdot 2\pi \\
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n=-2 & x=\frac{\pi }{3}+\left( -2 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -2 \right)\centerdot 2\pi \\
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n=2 & x=\frac{\pi }{3}+2\centerdot 2\pi & x=\frac{2\pi }{3}+2\centerdot 2\pi \\
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\end{array}</math>
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and so on.

Version vom 13:30, 30. Sep. 2008

We draw a unit circle and mark on those angles on the circle which have a \displaystyle y -coordinate of \displaystyle {\sqrt{3}}/{2}\;, in order to see which solutions lie between \displaystyle 0 and \displaystyle 2\pi .


In the first quadrant, we recognize \displaystyle x={\pi }/{3}\; as the angle which has a sine value of \displaystyle {\sqrt{3}}/{2}\; and then we have the reflectionally symmetric solution \displaystyle x=\pi -\frac{\pi }{3}=\frac{2\pi }{3} in the second quadrant.

Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of \displaystyle 2\pi


\displaystyle x=\frac{\pi }{3}+2n\pi and \displaystyle x=\frac{2\pi }{3}+2n\pi


where \displaystyle n is an arbitrary integer.

NOTE: when we write that the complete solution is given by


\displaystyle x=\frac{\pi }{3}+2n\pi and \displaystyle x=\frac{2\pi }{3}+2n\pi ,

this means that for every integer \displaystyle n, we obtain a solution to the equation:


\displaystyle \begin{array}{*{35}l} n=0 & x=\frac{\pi }{3} & x=\frac{2\pi }{3} \\ n=-1 & x=\frac{\pi }{3}+\left( -1 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -1 \right)\centerdot 2\pi \\ n=1 & x=\frac{\pi }{3}+1\centerdot 2\pi & x=\frac{2\pi }{3}+1\centerdot 2\pi \\ n=-2 & x=\frac{\pi }{3}+\left( -2 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -2 \right)\centerdot 2\pi \\ n=2 & x=\frac{\pi }{3}+2\centerdot 2\pi & x=\frac{2\pi }{3}+2\centerdot 2\pi \\ \end{array}

and so on.

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