Lösung 4.4:6a
Aus Online Mathematik Brückenkurs 1
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| - | {{ | + | If we move everything over to the left-hand side, |
| - | < | + | |
| - | {{ | + | |
| + | <math>\sin x\cos 3x-2\sin x=0</math> | ||
| + | |||
| + | |||
| + | we see that both terms have | ||
| + | <math>\text{sin }x\text{ }</math> | ||
| + | as a common factor which we can take out: | ||
| + | |||
| + | |||
| + | <math>\text{sin }x\text{ }\left( \cos 3x-2 \right)=0</math> | ||
| + | |||
| + | |||
| + | In this factorized version of the equation, we see the equation has a solution only when one of the factors | ||
| + | <math>\text{sin }x</math> | ||
| + | or | ||
| + | <math>\cos 3x-2</math> | ||
| + | is zero. The factor | ||
| + | <math>\text{sin }x</math> | ||
| + | is zero for all values of | ||
| + | <math>x</math> | ||
| + | that are given by | ||
| + | |||
| + | |||
| + | <math>x=n\pi </math> | ||
| + | ( | ||
| + | <math>n</math> | ||
| + | an arbitrary integer) | ||
| + | |||
| + | (see exercise 3.5:2c). The other factor | ||
| + | <math>\cos 3x-2</math> | ||
| + | can never be zero because the value of a cosine always lies between | ||
| + | <math>-\text{1 }</math> | ||
| + | and | ||
| + | <math>\text{1}</math>, which gives that the largest value of | ||
| + | <math>\cos 3x-2</math> | ||
| + | is | ||
| + | <math>-\text{1 }</math>. | ||
| + | |||
| + | The solutions are therefore | ||
| + | |||
| + | |||
| + | <math>x=n\pi </math> | ||
| + | ( | ||
| + | <math>n</math> | ||
| + | an arbitrary integer). | ||
Version vom 11:25, 1. Okt. 2008
If we move everything over to the left-hand side,
\displaystyle \sin x\cos 3x-2\sin x=0
we see that both terms have
\displaystyle \text{sin }x\text{ }
as a common factor which we can take out:
\displaystyle \text{sin }x\text{ }\left( \cos 3x-2 \right)=0
In this factorized version of the equation, we see the equation has a solution only when one of the factors
\displaystyle \text{sin }x
or
\displaystyle \cos 3x-2
is zero. The factor
\displaystyle \text{sin }x
is zero for all values of
\displaystyle x
that are given by
\displaystyle x=n\pi
(
\displaystyle n
an arbitrary integer)
(see exercise 3.5:2c). The other factor \displaystyle \cos 3x-2 can never be zero because the value of a cosine always lies between \displaystyle -\text{1 } and \displaystyle \text{1}, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -\text{1 }.
The solutions are therefore
\displaystyle x=n\pi
(
\displaystyle n
an arbitrary integer).
