Lösung 4.4:6c
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Version vom 15:17, 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.
