Lösung 4.4:8a

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If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
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{{Displayed math||<math>2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}</math>}}
Then, we see that we can take a factor <math>\cos x</math> out of both terms,
Then, we see that we can take a factor <math>\cos x</math> out of both terms,
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{{Displayed math||<math>\cos x\,(2\sin x-\sqrt{2}) = 0</math>}}
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{{Abgesetzte Formel||<math>\cos x\,(2\sin x-\sqrt{2}) = 0</math>}}
and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.
and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.
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This equation has the general solution
This equation has the general solution
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{{Displayed math||<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
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{{Abgesetzte Formel||<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
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<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
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{{Displayed math||<math>\left\{\begin{align}
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{{Abgesetzte Formel||<math>\left\{\begin{align}
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
x &= \frac{3\pi}{4}+2n\pi\,,
x &= \frac{3\pi}{4}+2n\pi\,,
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The complete solution of the equation is
The complete solution of the equation is
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{{Displayed math||<math>\left\{\begin{align}
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{{Abgesetzte Formel||<math>\left\{\begin{align}
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
x &= \frac{\pi}{2}+n\pi\,,\\[5pt]
x &= \frac{\pi}{2}+n\pi\,,\\[5pt]

Version vom 09:01, 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.

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