Lösung 4.4:2b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K (Lösning 4.4:2b moved to Solution 4.4:2b: Robot: moved page) |
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- | {{ | + | The equation |
- | < | + | <math>\cos x={1}/{2}\;</math> |
- | {{ | + | has the solution |
+ | <math>x={\pi }/{3}\;</math> | ||
+ | in the first quadrant, and the symmetric solution | ||
+ | <math>x={2\pi -\pi }/{3}\;={5\pi }/{3}\;</math> | ||
+ | in the fourth quadrant. | ||
+ | |||
[[Image:4_4_2_b.gif|center]] | [[Image:4_4_2_b.gif|center]] | ||
+ | |||
+ | Angle | ||
+ | <math>{\pi }/{3}\;</math> | ||
+ | Angle | ||
+ | <math>{5\pi }/{3}\;</math> | ||
+ | |||
+ | |||
+ | If we add multiples of | ||
+ | <math>2\pi </math> | ||
+ | to these two solutions, we obtain all the solutions | ||
+ | |||
+ | |||
+ | <math>x={\pi }/{3}\;+2n\pi </math> | ||
+ | and | ||
+ | <math>x={5\pi }/{3}\;+2n\pi </math> | ||
+ | |||
+ | |||
+ | where | ||
+ | <math>n</math> | ||
+ | is an arbitrary integer. |
Version vom 13:40, 30. Sep. 2008
The equation \displaystyle \cos x={1}/{2}\; has the solution \displaystyle x={\pi }/{3}\; in the first quadrant, and the symmetric solution \displaystyle x={2\pi -\pi }/{3}\;={5\pi }/{3}\; in the fourth quadrant.
Angle \displaystyle {\pi }/{3}\; Angle \displaystyle {5\pi }/{3}\;
If we add multiples of
\displaystyle 2\pi
to these two solutions, we obtain all the solutions
\displaystyle x={\pi }/{3}\;+2n\pi
and
\displaystyle x={5\pi }/{3}\;+2n\pi
where
\displaystyle n
is an arbitrary integer.