Lösung 4.4:3a
Aus Online Mathematik Brückenkurs 1
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- | The right-hand side of the equation is a constant, so the equation is in fact a normal trigonometric equation of the type | + | The right-hand side of the equation is a constant, so the equation is in fact a normal trigonometric equation of the type <math>\cos x = a\,</math>. |
- | <math>\ | + | |
- | In this case, we can see directly that one solution is | + | In this case, we can see directly that one solution is <math>x = \pi/6\,</math>. Using the unit circle, it follows that <math>x = 2\pi - \pi/6 = 11\pi/6\,</math> is the only other solution between <math>0</math> and <math>2\pi\,</math>. |
- | <math>x= | + | |
- | <math>x=2\pi - | + | |
- | is the only other solution between | + | |
- | <math>0</math> | + | |
- | and | + | |
- | <math> | + | |
[[Image:4_4_3_a.gif|center]] | [[Image:4_4_3_a.gif|center]] | ||
- | We obtain all solutions to the equation if we add multiples of | + | We obtain all solutions to the equation if we add multiples of <math>2\pi</math> to the two solutions above, |
- | <math> | + | |
- | to the two solutions above | + | |
+ | {{Displayed math||<math>x = \frac{\pi}{6} + 2n\pi\qquad\text{and}\qquad x = \frac{11\pi}{6} + 2n\pi\,,</math>}} | ||
- | + | where ''n'' is an arbitrary integer. | |
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- | where | + | |
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- | is an arbitrary integer. | + |
Version vom 12:49, 13. Okt. 2008
The right-hand side of the equation is a constant, so the equation is in fact a normal trigonometric equation of the type \displaystyle \cos x = a\,.
In this case, we can see directly that one solution is \displaystyle x = \pi/6\,. Using the unit circle, it follows that \displaystyle x = 2\pi - \pi/6 = 11\pi/6\, is the only other solution between \displaystyle 0 and \displaystyle 2\pi\,.
We obtain all solutions to the equation if we add multiples of \displaystyle 2\pi to the two solutions above,
where n is an arbitrary integer.