Lösung 4.4:2a

Aus Online Mathematik Brückenkurs 1

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We draw a unit circle and mark on those angles on the circle which have a
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We draw a unit circle and mark those angles on the circle which have a ''y''-coordinate of <math>\sqrt{3}/2</math>, in order to see which solutions lie between
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<math>y</math>
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<math>0</math> and <math>2\pi</math>.
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-coordinate of
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<math>{\sqrt{3}}/{2}\;</math>, in order to see which solutions lie between
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<math>0</math>
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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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In the first quadrant, we recognize <math>x = \pi/3</math> as the angle which has a sine value of <math>\sqrt{3}/2</math> and then we have the reflectionally symmetric solution <math>x = \pi - \pi/3 = 2\pi/3</math> in the second quadrant.
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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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Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of <math>2\pi</math>
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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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{{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.
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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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Note: When we write that the complete solution is given by
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<math>n</math>, we obtain a solution to the equation:
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{{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,}</math>}}
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<math>\begin{array}{*{35}l}
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this means that for every integer ''n'', we obtain a solution to the equation:
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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.
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{{Displayed math||<math>\begin{array}{llll}
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&n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt]
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&n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt]
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&n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt]
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&n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt]
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&n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt]
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&\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots
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\end{array}</math>}}

Version vom 14:13, 10. Okt. 2008

We draw a unit circle and mark those angles on the circle which have a 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 - \pi/3 = 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

Vorlage:Displayed math

where n is an arbitrary integer.


Note: When we write that the complete solution is given by

Vorlage:Displayed math

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

Vorlage:Displayed math

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