Lösung 4.4:6c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
|||
| Zeile 1: | Zeile 1: | ||
| - | If we use the trigonometric relation | + | If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as |
| - | <math>\ | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<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 | ||
| - | + | {{Displayed math||<math>\sin u = \sin v</math>}} | |
| - | <math>\sin u=\sin v | + | |
| - | + | ||
is satisfied if | is satisfied if | ||
| + | {{Displayed math||<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 | + | |
| - | + | ||
| - | is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}} | ||
i.e. | i.e. | ||
| + | {{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}} | ||
| - | + | The solutions to the equation are thus | |
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>\left\{\begin{align} | ||
| + | x &= \frac{2n\pi}{3}\,,\\[5pt] | ||
| + | x &= \pi + 2n\pi\,, | ||
| + | \end{align}\right.</math>}} | ||
| - | + | where ''n'' is an arbitrary integer. | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | an arbitrary integer | + | |
Version vom 06:55, 14. Okt. 2008
If we use the trigonometric relation \displaystyle \sin (-x) = -\sin x, the equation can be rewritten as
In exercise 4.4:5a, we saw that an equality of the type
is satisfied if
where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
i.e.
The solutions to the equation are thus
where n is an arbitrary integer.
