Lösung 4.4:6a

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If we move everything over to the left-hand side,
If we move everything over to the left-hand side,
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{{Displayed math||<math>\sin x\cos 3x-2\sin x=0</math>}}
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{{Abgesetzte Formel||<math>\sin x\cos 3x-2\sin x=0</math>}}
we see that both terms have <math>\sin x</math> as a common factor which we can take out,
we see that both terms have <math>\sin x</math> as a common factor which we can take out,
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{{Displayed math||<math>\sin x (\cos 3x-2) = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin x (\cos 3x-2) = 0\,\textrm{.}</math>}}
In this factorized version of the equation, we see the equation has a solution only when one of the factors <math>\sin x</math> or <math>\cos 3x-2</math> is zero. The factor <math>\sin x</math> is zero for all values of ''x'' that are given by
In this factorized version of the equation, we see the equation has a solution only when one of the factors <math>\sin x</math> or <math>\cos 3x-2</math> is zero. The factor <math>\sin x</math> is zero for all values of ''x'' that are given by
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{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
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{{Abgesetzte Formel||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
(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>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.
(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>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.
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The solutions are therefore
The solutions are therefore
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{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
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{{Abgesetzte Formel||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}

Version vom 09:00, 22. 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 \sin x as a common factor which we can take out,

\displaystyle \sin x (\cos 3x-2) = 0\,\textrm{.}

In this factorized version of the equation, we see the equation has a solution only when one of the factors \displaystyle \sin x or \displaystyle \cos 3x-2 is zero. The factor \displaystyle \sin x is zero for all values of x that are given by

\displaystyle x=n\pi\qquad\text{(n is 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 -1 and \displaystyle 1, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -1.

The solutions are therefore

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer).}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:6a“