Lösung 4.4:4

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The idea is first to find the general solution to the equation and then to see which angles lie between
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<center> [[Image:4_4_4-1(3).gif]] </center>
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<math>0^{\circ }</math>
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and
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<math>360^{\circ }</math>.
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<center> [[Image:4_4_4-2(3).gif]] </center>
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If we start by considering the expression
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<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
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<center> [[Image:4_4_4-3(3).gif]] </center>
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as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
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<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>
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There is then a further solution which satisfies
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<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
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<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
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lies in the third quadrant and makes the same angle with the negative y-axis as
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<math>\text{1}00^{\circ }</math>
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makes with the positive
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<math>y</math>
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-axis, i.e.
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<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
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makes an angle
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<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
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with the negative
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<math>y</math>
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-axis and consequently
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<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
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[[Image:4_4_4.gif|center]]
[[Image:4_4_4.gif|center]]
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 +
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There is then a further solution which satisfies
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<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
lies in the third quadrant and makes the same angle with the negative y-axis as
 +
<math>\text{1}00^{\circ }</math>
 +
makes with the positive
 +
<math>y</math>
 +
-axis, i.e.
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
makes an angle
 +
<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
 +
with the negative
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<math>y</math>
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-axis and consequently
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 +
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<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
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 +
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FIGURE1 FIGURE2
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Now it is easy to write down the general solution,
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<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>
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and
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<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>
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and if we make
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<math>v</math>
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the subject, we get
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<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>
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and
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<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>
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EQ6
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For different values of the integers
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<math>n</math>, we see that the corresponding solutions are:
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<math>\begin{array}{*{35}l}
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\cdots \cdots & \cdots \cdots & \cdots \cdots \\
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n=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\
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n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\
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n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\
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n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\
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n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\
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n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\
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\cdots \cdots & \cdots \cdots & \cdots \cdots \\
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\end{array}</math>
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From the table, we see that the solutions that are between
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<math>0^{\circ }</math>
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and
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<math>360^{\circ }</math>
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are
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<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }</math>
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and
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<math>v=300^{\circ }</math>

Version vom 10:20, 1. Okt. 2008

The idea is first to find the general solution to the equation and then to see which angles lie between \displaystyle 0^{\circ } and \displaystyle 360^{\circ }.

If we start by considering the expression \displaystyle \text{2}v+\text{1}0^{\circ }\text{ } as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is


\displaystyle \text{2}v+\text{1}0^{\circ }=110^{\circ }


There is then a further solution which satisfies \displaystyle 0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }, where \displaystyle \text{2}v+\text{1}0^{\circ }\text{ } lies in the third quadrant and makes the same angle with the negative y-axis as \displaystyle \text{1}00^{\circ } makes with the positive \displaystyle y -axis, i.e. \displaystyle \text{2}v+\text{1}0^{\circ }\text{ } makes an angle \displaystyle \text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ } with the negative \displaystyle y -axis and consequently


\displaystyle \text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }



There is then a further solution which satisfies \displaystyle 0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }, where \displaystyle \text{2}v+\text{1}0^{\circ }\text{ } lies in the third quadrant and makes the same angle with the negative y-axis as \displaystyle \text{1}00^{\circ } makes with the positive \displaystyle y -axis, i.e. \displaystyle \text{2}v+\text{1}0^{\circ }\text{ } makes an angle \displaystyle \text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ } with the negative \displaystyle y -axis and consequently


\displaystyle \text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }


FIGURE1 FIGURE2

Now it is easy to write down the general solution,


\displaystyle \text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ } and

\displaystyle \text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }


and if we make \displaystyle v the subject, we get


\displaystyle v=50^{\circ }+n\centerdot 180^{\circ } and

\displaystyle v=120^{\circ }+n\centerdot 180^{\circ } EQ6

For different values of the integers \displaystyle n, we see that the corresponding solutions are:


\displaystyle \begin{array}{*{35}l} \cdots \cdots & \cdots \cdots & \cdots \cdots \\ n=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\ n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\ n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\ n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\ n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\ n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\ \cdots \cdots & \cdots \cdots & \cdots \cdots \\ \end{array}


From the table, we see that the solutions that are between \displaystyle 0^{\circ } and \displaystyle 360^{\circ } are


\displaystyle v=50,\quad v=120^{\circ },\quad v=230^{\circ } and \displaystyle v=300^{\circ }

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