Lösung 4.4:8a
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Version vom 15:19, 22. Okt. 2008
If we use the formula for double angles, \displaystyle \sin 2x = 2\sin x\cos x, and move all the terms over to the left-hand side, the equation becomes
| \displaystyle 2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.} |
Then, we see that we can take a factor \displaystyle \cos x out of both terms,
| \displaystyle \cos x\,(2\sin x-\sqrt{2}) = 0 |
and hence divide up the equation into two cases. The equation is satisfied either if \displaystyle \cos x = 0 or if \displaystyle 2\sin x-\sqrt{2} = 0\,.
\displaystyle \cos x = 0:
This equation has the general solution
| \displaystyle x = \frac{\pi}{2}+n\pi\qquad(n is an arbitrary integer). |
\displaystyle 2\sin x-\sqrt{2}=0:
If we collect \displaystyle \sin x on the left-hand side, we obtain the equation \displaystyle \sin x = 1/\!\sqrt{2}, which has the general solution
| \displaystyle \left\{\begin{align}
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right. |
where n is an arbitrary integer.
The complete solution of the equation is
| \displaystyle \left\{\begin{align}
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{\pi}{2}+n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right. |
where n is an arbitrary integer.
