Lösung 4.4:6b

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After moving the terms over to the left-hand side, so that
After moving the terms over to the left-hand side, so that
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{{Displayed math||<math>\sqrt{2}\sin x\cos x-\cos x=0</math>}}
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{{Abgesetzte Formel||<math>\sqrt{2}\sin x\cos x-\cos x=0</math>}}
we see that we can take out a common factor <math>\cos x</math>,
we see that we can take out a common factor <math>\cos x</math>,
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{{Displayed math||<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}
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{{Abgesetzte Formel||<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}
and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
Zeile 14: Zeile 14:
This basic equation has solutions <math>x=\pi/2</math> and <math>x=3\pi/2</math> in the unit circle, and from this we see that the general solution is
This basic equation has solutions <math>x=\pi/2</math> and <math>x=3\pi/2</math> in the unit circle, and from this we see that the general solution is
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{{Displayed math||<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
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{{Abgesetzte Formel||<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
where ''n'' is an arbitrary integer. Because the angles <math>\pi/2</math> and
where ''n'' is an arbitrary integer. Because the angles <math>\pi/2</math> and
<math>3\pi/2</math> differ by <math>\pi</math>, the solutions can be summarized as
<math>3\pi/2</math> differ by <math>\pi</math>, the solutions can be summarized as
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{{Displayed math||<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.
Zeile 28: Zeile 28:
If we rearrange the equation, we obtain the basic equation as <math>\sin x = 1/\!\sqrt{2}</math>, which has the solutions <math>x=\pi/4</math> and <math>x=3\pi /4</math> in the unit circle and hence the general solution
If we rearrange the equation, we obtain the basic equation as <math>\sin x = 1/\!\sqrt{2}</math>, which has the solutions <math>x=\pi/4</math> and <math>x=3\pi /4</math> in the unit circle and hence the general solution
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{{Displayed math||<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.
Zeile 35: Zeile 35:
All in all, the original equation has the solutions
All in all, the original equation has the solutions
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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 09:00, 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.