Lösung 4.4:6b
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Version vom 15:17, 22. Okt. 2008
After moving the terms over to the left-hand side, so that
| \displaystyle \sqrt{2}\sin x\cos x-\cos x=0 |
we see that we can take out a common factor \displaystyle \cos x,
| \displaystyle \cos x (\sqrt{2}\sin x-1) = 0 |
and that the equation is only satisfied if at least one of the factors \displaystyle \cos x or \displaystyle \sqrt{2}\sin x - 1 is zero. Thus, there are two cases:
\displaystyle \cos x=0:
This basic equation has solutions \displaystyle x=\pi/2 and \displaystyle x=3\pi/2 in the unit circle, and from this we see that the general solution is
| \displaystyle x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,, |
where n is an arbitrary integer. Because the angles \displaystyle \pi/2 and \displaystyle 3\pi/2 differ by \displaystyle \pi, the solutions can be summarized as
| \displaystyle x=\frac{\pi}{2}+n\pi\,, |
where n is an arbitrary integer.
\displaystyle \sqrt{2}\sin x - 1 = 0:
If we rearrange the equation, we obtain the basic equation as \displaystyle \sin x = 1/\!\sqrt{2}, which has the solutions \displaystyle x=\pi/4 and \displaystyle x=3\pi /4 in the unit circle and hence the general solution
| \displaystyle x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,, |
where n is an arbitrary integer.
All in all, the original equation has the solutions
| \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.
