Lösung 4.4:6b
Aus Online Mathematik Brückenkurs 1
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
Zeile 1: | Zeile 1: | ||
After moving the terms over to the left-hand side, so that | After moving the terms over to the left-hand side, so that | ||
- | {{ | + | {{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>, | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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.