Lösung 4.4:6c

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If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as
If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as
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{{Displayed math||<math>\sin 2x = \sin (-x)\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin 2x = \sin (-x)\,\textrm{.}</math>}}
In exercise 4.4:5a, we saw that an equality of the type
In exercise 4.4:5a, we saw that an equality of the type
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{{Displayed math||<math>\sin u = \sin v</math>}}
+
{{Abgesetzte Formel||<math>\sin u = \sin v</math>}}
is satisfied if
is satisfied if
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{{Displayed math||<math>u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,</math>}}
where ''n'' is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
where ''n'' is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
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{{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}}
i.e.
i.e.
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{{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}}
The solutions to the equation are thus
The solutions to the equation are thus
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{{Displayed math||<math>\left\{\begin{align}
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{{Abgesetzte Formel||<math>\left\{\begin{align}
x &= \frac{2n\pi}{3}\,,\\[5pt]
x &= \frac{2n\pi}{3}\,,\\[5pt]
x &= \pi + 2n\pi\,,
x &= \pi + 2n\pi\,,

Version vom 09:00, 22. Okt. 2008

If we use the trigonometric relation \displaystyle \sin (-x) = -\sin x, the equation can be rewritten as

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

In exercise 4.4:5a, we saw that an equality of the type

\displaystyle \sin u = \sin v

is satisfied if

\displaystyle u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

\displaystyle 2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,

i.e.

\displaystyle 3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}

The solutions to the equation are thus

\displaystyle \left\{\begin{align}

x &= \frac{2n\pi}{3}\,,\\[5pt] x &= \pi + 2n\pi\,, \end{align}\right.

where n is an arbitrary integer.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:6c“