Lösung 4.4:3c
Aus Online Mathematik Brückenkurs 1
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It is then easy to set up the general solution by adding multiples of <math>360^{\circ}\,</math>, | It is then easy to set up the general solution by adding multiples of <math>360^{\circ}\,</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x + 40^{\circ} = 65^{\circ} + n\cdot 360^{\circ}\qquad\text{and}\qquad x + 40^{\circ} = 115^{\circ} + n\cdot 360^{\circ}</math>}} |
for all integers ''n'', which gives | for all integers ''n'', which gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x = 25^{\circ} + n\cdot 360^{\circ}\qquad\text{and}\qquad x=75^{\circ} + n\cdot 360^{\circ}\,\textrm{.}</math>}} |
Version vom 08:59, 22. Okt. 2008
If we consider the entire expression \displaystyle x + 40^{\circ} as an unknown, we have a basic trigonometric equation and can, with the aid of the unit circle, see that there are two solutions to the equation for \displaystyle 0^{\circ}\le x+40^{\circ}\le 360^{\circ} namely \displaystyle x+40^{\circ} = 65^{\circ} and the symmetric solution \displaystyle x + 40^{\circ} = 180^{\circ} - 65^{\circ} = 115^{\circ}\,.
It is then easy to set up the general solution by adding multiples of \displaystyle 360^{\circ}\,,
| \displaystyle x + 40^{\circ} = 65^{\circ} + n\cdot 360^{\circ}\qquad\text{and}\qquad x + 40^{\circ} = 115^{\circ} + n\cdot 360^{\circ} |
for all integers n, which gives
| \displaystyle x = 25^{\circ} + n\cdot 360^{\circ}\qquad\text{and}\qquad x=75^{\circ} + n\cdot 360^{\circ}\,\textrm{.} |
