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 <math>0^{\circ}</math> and <math>360^{\circ}\,</math>.
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Wir finden zuerst die allgemeine Lösung der Gleichung, und sehen danach welche der Winkeln im Intervall zwischen <math>0^{\circ}</math> und <math>360^{\circ}\,</math> liegen.
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If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
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Wir betrachten zuerst den Ausdruck <math>2v+10^{\circ}</math>, und erhalten die Lösung
{{Abgesetzte Formel||<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}
{{Abgesetzte Formel||<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}
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Es gibt noch eine Lösung im Einheitskreis, nämlich im dritten Quadrant, mit denselben Winkel zur negativen ''y''-Achse, wie
There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
with the negative ''y''-axis and consequently
with the negative ''y''-axis and consequently

Version vom 09:30, 7. Apr. 2009

Wir finden zuerst die allgemeine Lösung der Gleichung, und sehen danach welche der Winkeln im Intervall zwischen \displaystyle 0^{\circ} und \displaystyle 360^{\circ}\, liegen.

Wir betrachten zuerst den Ausdruck \displaystyle 2v+10^{\circ}, und erhalten die Lösung

\displaystyle 2v + 10^{\circ} = 110^{\circ}\,\textrm{.}

Es gibt noch eine Lösung im Einheitskreis, nämlich im dritten Quadrant, mit denselben Winkel zur negativen y-Achse, wie There is then a further solution which satisfies \displaystyle 0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}, where \displaystyle 2v+10^{\circ} lies in the third quadrant and makes the same angle with the negative y-axis as \displaystyle 100^{\circ} makes with the positive y-axis, i.e. \displaystyle 2v + 10^{\circ} makes an angle \displaystyle 110^{\circ} - 90^{\circ} = 20^{\circ} with the negative y-axis and consequently

\displaystyle 2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}

Now it is easy to write down the general solution,

\displaystyle \left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.

and if we make v the subject, we get

\displaystyle \left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.

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


\displaystyle \cdots\cdots \displaystyle \cdots\cdots \displaystyle \cdots\cdots
\displaystyle n=-2:   \displaystyle v = 50^{\circ} - 2\cdot 180^{\circ} = -310^{\circ}   \displaystyle v = 120^{\circ } - 2\cdot 180^{\circ} = -240^{\circ}
\displaystyle n=-1: \displaystyle v = 50^{\circ} - 1\cdot 180^{\circ} = -130^{\circ} \displaystyle v = 120^{\circ} - 1\cdot 180^{\circ} = -60^{\circ}
\displaystyle n=0: \displaystyle v = 50^{\circ} + 0\cdot 180^{\circ} = 50^{\circ} \displaystyle v = 120^{\circ} + 0\cdot 180^{\circ} = 120^{\circ}
\displaystyle n=1: \displaystyle v = 50^{\circ} + 1\cdot 180^{\circ} = 230^{\circ} \displaystyle v = 120^{\circ} + 1\cdot 180^{\circ} = 300^{\circ}
\displaystyle n=2: \displaystyle v = 50^{\circ} + 2\cdot 180^{\circ} = 410^{\circ} \displaystyle v = 120^{\circ} + 2\cdot 180^{\circ} = 480^{\circ}
\displaystyle n=3: \displaystyle v = 50^{\circ} + 3\cdot 180^{\circ} = 590^{\circ} \displaystyle v = 120^{\circ} + 3\cdot 180^{\circ} = 660^{\circ}
\displaystyle \cdots\cdots \displaystyle \cdots\cdots \displaystyle \cdots\cdots


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

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