Lösung 4.4:7a - Versionsgeschichte

Moni: Wortwahl

2009-08-07T22:49:07Z

Wortwahl

← Nächstältere Version		Version vom 22:49, 7. Aug. 2009
Zeile 21:		Zeile 21:
	{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{und}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{und}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}

-	Nachdem <math>t=\sin x</math>, muss ''x'' der Gleichungen <math>\sin x = \tfrac{1}{2}</math> oder <math>\sin x = -1\,</math> ~~genügen~~.	+	Nachdem <math>t=\sin x</math>, muss ''x'' der Gleichungen <math>\sin x = \tfrac{1}{2}</math> oder <math>\sin x = -1\,</math> erfüllen.

Markus.bez: Sprache und Formulierung

2009-06-19T14:54:51Z

Sprache und Formulierung

Martin um 11:54, 7. Apr. 2009

2009-04-07T11:54:32Z

← Nächstältere Version		Version vom 11:54, 7. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If we examine the equation, we see that~~ <math>x</math> ~~only occurs as~~ <math>\sin x</math> ~~and it can therefore be appropriate to take an intermediary step and solve for~~ <math>\sin x</math>, ~~instead of trying to solve for~~ <math>x</math> ~~directly.~~	+	Wir sehen dass <math>x</math> nur in den <math>\sin x</math>-Termen auftritt. Daher Lösen wir zuerst die Gleichung für <math>\sin x</math>, statt direkt für <math>x</math>

-	~~If we write~~ <math>t = \sin x</math> ~~and treat <math>t</math> as a new unknown variable, the equation becomes~~	+	Wir benennen <math>t = \sin x</math> und erhalten die Gleichung

	{{Abgesetzte Formel\|\|<math>2t^2 + t = 1</math>}}		{{Abgesetzte Formel\|\|<math>2t^2 + t = 1</math>}}

-	~~and it is expressed completely in terms of~~ <math>t</math>. ~~This is a normal quadratic equation; after dividing by~~ 2~~, we complete the square on the left-hand side,~~	+	Dies ist eine quadratische Gleichung für <math>t</math>. Nach division durch 2 erhalten wir die Gleichung

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	~~and then obtain the equation~~	+	oder

	{{Abgesetzte Formel\|\|<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}		{{Abgesetzte Formel\|\|<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}

-	~~which has the solutions~~ <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, ~~i.e.~~	+	mit den Lösungen <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, also

-	{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{~~and~~}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{und}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}

-		+	Nachdem <math>t=\sin x</math>, müssen die Lösungen ''x'', der Gleichung, einer der Gleichungen <math>\sin x = \tfrac{1}{2}</math> oder <math>\sin x = -1\,</math> erfüllen.
-	~~Because~~ <math>t=\sin x</math>, ~~this means that the values of~~ ''x'' ~~that satisfy the equation in the exercise will necessarily satisfy one of the basic equations~~ <math>\sin x = \tfrac{1}{2}</math> or <math>\sin x = -1\,</math>.	+


	<math>\sin x = \frac{1}{2}</math>:		<math>\sin x = \frac{1}{2}</math>:

-	~~This equation has the solutions~~ <math>x = \pi/6</math> ~~and~~ <math>x = \pi - \pi/6 = 5\pi/6</math> ~~in the unit circle and the general solution is~~	+	Diese Gleichung hat die Lösungen <math>x = \pi/6</math> und <math>x = \pi - \pi/6 = 5\pi/6</math> im Einheitskreis, und daher die allgemeinen Lösungen

-	{{Abgesetzte Formel\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{~~and~~}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{und}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}
-		+
-	~~where ''n'' is an arbitrary integer.~~	+


	<math>\sin x = -1</math>:		<math>\sin x = -1</math>:

-	~~The equation has only one solution~~ <math>x = 3\pi/2</math> ~~in the unit circle~~, ~~and the general solution is therefore~~	+	Hat nur die Lösung <math>x = 3\pi/2</math> im Einheitskreis, und daher die allgemeinen Lösungen

	{{Abgesetzte Formel\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}		{{Abgesetzte Formel\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}

-	where ''n'' is an arbitrary integer.


-	~~All of the solution to the equation are given by~~	+	Zusammen erhalten wir also die Lösungen

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 50:		Zeile 46:
	x &= 3\pi/2+2n\pi\,,		x &= 3\pi/2+2n\pi\,,
	\end{align}\right.</math>}}		\end{align}\right.</math>}}
-
-	where ''n'' is an arbitrary integer.

Tekbot: hat „Solution 4.4:7a“ nach „Lösung 4.4:7a“ verschoben: Robot: moved page

2008-10-22T15:18:04Z

hat „Solution 4.4:7a“ nach „Lösung 4.4:7a“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:18, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2008-10-22T09:00:48Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 09:00, 22. Okt. 2008
Zeile 3:		Zeile 3:
	If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes		If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes

-	{{~~Displayed math~~\|\|<math>2t^2 + t = 1</math>}}	+	{{Abgesetzte Formel\|\|<math>2t^2 + t = 1</math>}}

	and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,		and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	t^2 + \frac{1}{2}t - \frac{1}{2}		t^2 + \frac{1}{2}t - \frac{1}{2}
	&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]		&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
Zeile 15:		Zeile 15:
	and then obtain the equation		and then obtain the equation

-	{{~~Displayed math~~\|\|<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}	+	{{Abgesetzte Formel\|\|<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}

	which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.		which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.

-	{{~~Displayed math~~\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}


Zeile 29:		Zeile 29:
	This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is		This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is

-	{{~~Displayed math~~\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}

	where ''n'' is an arbitrary integer.		where ''n'' is an arbitrary integer.
Zeile 38:		Zeile 38:
	The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore		The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore

-	{{~~Displayed math~~\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}

	where ''n'' is an arbitrary integer.		where ''n'' is an arbitrary integer.
Zeile 45:		Zeile 45:
	All of the solution to the equation are given by		All of the solution to the equation are given by

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
	x &= \pi/6+2n\pi\,,\\[5pt]		x &= \pi/6+2n\pi\,,\\[5pt]
	x &= 5\pi/6+2n\pi\,,\\[5pt]		x &= 5\pi/6+2n\pi\,,\\[5pt]

Tek um 07:32, 14. Okt. 2008

2008-10-14T07:32:07Z

← Nächstältere Version		Version vom 07:32, 14. Okt. 2008
Zeile 1:		Zeile 1:
-	If we examine the equation, we see that	+	If we examine the equation, we see that <math>x</math> only occurs as <math>\sin x</math> and it can therefore be appropriate to take an intermediary step and solve for <math>\sin x</math>, instead of trying to solve for <math>x</math> directly.
-	<math>x</math>	+
-	only occurs as	+
-	<math>\~~text{~~sin }x~~\text{ }~~</math>	+
-	and it can therefore be appropriate to take an intermediary step and solve for	+
-	<math>\~~text{~~sin }x</math>, instead of trying to solve for	+
-	<math>x</math>	+
-	directly.	+

-	If we write	+	If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
-	<math>t=\sin x</math>	+
-	and treat	+
-	<math>t</math>	+
-	as a new unknown variable, the equation becomes	+

		+	{{Displayed math\|\|<math>2t^2 + t = 1</math>}}

-	~~<math>2t^{2}+t=1</math>~~	+	and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,
-		+
-		+
-	~~when~~ it is expressed completely in terms of	+
-	<math>t</math>. This is a normal ~~second-degree~~ equation; after dividing by	+
-	~~<math>\text{~~2~~}</math>~~, we complete the square on the left-hand side,	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\~~	+
-	~~& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	t^2 + \frac{1}{2}t - \frac{1}{2}
		+	&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
		+	&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16}
		+	\end{align}</math>}}

	and then obtain the equation		and then obtain the equation

		+	{{Displayed math\|\|<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}

-	~~<math>\left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}</math>~~	+	which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.
-		+
-		+
-	which has the solutions	+
-	<math>t=-\~~frac~~{1}{4}\pm \sqrt{\~~frac~~{9}{16}}=-\~~frac~~{1}{4}\pm \~~frac~~{3}{4}</math>,	+
-	i.e.	+
-	~~<math>t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}</math>~~	+
-	~~and~~	+
-	~~<math>t=-\frac{1}{4}-\frac{3}{4}=-1</math>~~	+
-		+

-	~~Because~~	+	{{Displayed math\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}
-	<math>t=\~~sin x</math>, this means that the values of~~	+
-	~~<math>x</math>~~	+
-	~~that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,~~	+
-	~~<math>~~\~~text~~{~~sin~~ }x=\frac{1}{2}~~</math>~~	+
-	or	+
-	~~<math>~~\text{~~sin~~ }x=-\~~text~~{1}.</math>	+


		+	Because <math>t=\sin x</math>, this means that the values of ''x'' that satisfy the equation in the exercise will necessarily satisfy one of the basic equations <math>\sin x = \tfrac{1}{2}</math> or <math>\sin x = -1\,</math>.

-	<math>\text{sin }x=\frac{1}{2}</math>: this equation has the solutions
-	<math>x={\pi }/{6}\;</math>
-	and
-	<math>x=\pi -{\pi }/{6}\;=5{\pi }/{6}\;</math>
-	in the unit circle and the general solution is

		+	<math>\sin x = \frac{1}{2}</math>:

-	<math>x=~~\frac{~~\pi }{6~~}+2n\pi~~ </math>	+	This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is
-	and	+
-	<math>x=\~~frac{5~~\pi }{6~~}+2n~~\pi </math>	+

		+	{{Displayed math\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}

-	where	+	where ''n'' is an arbitrary integer.
-	~~<math>~~n~~\text{ }</math>~~	+
-	is an arbitrary integer.	+


-	<math>\~~text{~~sin }x=-~~\text{~~1}</math>: ~~the equation has only one solution~~	+	<math>\sin x = -1</math>:
-	~~<math>x={3\pi }/{2}\;</math>~~	+
-	~~in the unit circle, and the general solution is therefore~~	+

		+	The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore

-	<math>x=\frac{3\pi }{2}+2n\pi </math>	+	{{Displayed math\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}

		+	where ''n'' is an arbitrary integer.

-	where
-	<math>n\text{ }</math>
-	is an arbitrary integer.

	All of the solution to the equation are given by		All of the solution to the equation are given by

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	x &= \pi/6+2n\pi\,,\\[5pt]
		+	x &= 5\pi/6+2n\pi\,,\\[5pt]
		+	x &= 3\pi/2+2n\pi\,,
		+	\end{align}\right.</math>}}

-	~~<math>\left\{ \begin{array}{*{35}l}~~	+	where ''n'' is an arbitrary integer.
-	~~x={\pi }/{6}\;+2n\pi \\~~	+
-	~~x={5\pi }/{6}\;+2n\pi \\~~	+
-	~~x={3\pi }/{2}\;+2n\pi \\~~	+
-	~~\end{array} \right.</math>~~	+
-	(	+
-	~~<math>~~n~~\text{ }</math>~~	+
-	an arbitrary integer)	+

Ian um 12:49, 1. Okt. 2008

2008-10-01T12:49:28Z

← Nächstältere Version		Version vom 12:49, 1. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	If we examine the equation, we see that
-	<~~center~~> ~~[[Image:4_4_7a~~-1(2).~~gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	only occurs as
-	{{~~NAVCONTENT_START~~}}	+	<math>\text{sin }x\text{ }</math>
-	<~~center~~> ~~[[Image~~:~~4_4_7a~~-2(2~~).gif]]~~ </~~center~~>	+	and it can therefore be appropriate to take an intermediary step and solve for
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\text{sin }x</math>, instead of trying to solve for
		+	<math>x</math>
		+	directly.
		+
		+	If we write
		+	<math>t=\sin x</math>
		+	and treat
		+	<math>t</math>
		+	as a new unknown variable, the equation becomes
		+
		+
		+	<math>2t^{2}+t=1</math>
		+
		+
		+	when it is expressed completely in terms of
		+	<math>t</math>. This is a normal second-degree equation; after dividing by
		+	<math>\text{2}</math>, we complete the square on the left-hand side,
		+
		+
		+	<math>\begin{align}
		+	& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\
		+	& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\
		+	\end{align}</math>
		+
		+
		+	and then obtain the equation
		+
		+
		+	<math>\left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}</math>
		+
		+
		+	which has the solutions
		+	<math>t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}</math>,
		+	i.e.
		+	<math>t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}</math>
		+	and
		+	<math>t=-\frac{1}{4}-\frac{3}{4}=-1</math>
		+
		+
		+	Because
		+	<math>t=\sin x</math>, this means that the values of
		+	<math>x</math>
		+	that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,
		+	<math>\text{sin }x=\frac{1}{2}</math>
		+	or
		+	<math>\text{sin }x=-\text{1}.</math>
		+
		+
		+
		+	<math>\text{sin }x=\frac{1}{2}</math>: this equation has the solutions
		+	<math>x={\pi }/{6}\;</math>
		+	and
		+	<math>x=\pi -{\pi }/{6}\;=5{\pi }/{6}\;</math>
		+	in the unit circle and the general solution is
		+
		+
		+	<math>x=\frac{\pi }{6}+2n\pi </math>
		+	and
		+	<math>x=\frac{5\pi }{6}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer.
		+
		+
		+	<math>\text{sin }x=-\text{1}</math>: the equation has only one solution
		+	<math>x={3\pi }/{2}\;</math>
		+	in the unit circle, and the general solution is therefore
		+
		+
		+	<math>x=\frac{3\pi }{2}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer.
		+
		+	All of the solution to the equation are given by
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x={\pi }/{6}\;+2n\pi \\
		+	x={5\pi }/{6}\;+2n\pi \\
		+	x={3\pi }/{2}\;+2n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n\text{ }</math>
		+	an arbitrary integer)

Tekbot: Lösning 4.4:7a moved to Solution 4.4:7a: Robot: moved page

2008-09-09T10:07:43Z

Lösning 4.4:7a moved to Solution 4.4:7a: Robot: moved page

← Nächstältere Version

Version vom 10:07, 9. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T07:20:59Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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2008-04-03T12:41:44Z

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← Nächstältere Version		Version vom 14:54, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir sehen dass <math>x</math> nur in den <math>\sin x</math>-Termen auftritt. Daher Lösen wir zuerst die Gleichung für <math>\sin x</math>, statt direkt für <math>x</math>	+	Wir sehen, dass <math>x</math> nur in den <math>\sin x</math>-Termen auftritt. Daher Lösen wir zuerst die Gleichung für <math>\sin x</math>, statt direkt für <math>x</math>

	Wir benennen <math>t = \sin x</math> und erhalten die Gleichung		Wir benennen <math>t = \sin x</math> und erhalten die Gleichung
Zeile 5:		Zeile 5:
	{{Abgesetzte Formel\|\|<math>2t^2 + t = 1</math>}}		{{Abgesetzte Formel\|\|<math>2t^2 + t = 1</math>}}

-	Dies ist eine quadratische Gleichung für <math>t</math>. Nach ~~division~~ durch 2 erhalten wir die Gleichung	+	Dies ist eine quadratische Gleichung für <math>t</math>. Nach Division durch 2 erhalten wir die Gleichung

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 21:		Zeile 21:
	{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{und}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{und}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}

-	Nachdem <math>t=\sin x</math>, ~~müssen die Lösungen~~ ''x''~~, der Gleichung, einer~~ der Gleichungen <math>\sin x = \tfrac{1}{2}</math> oder <math>\sin x = -1\,</math> ~~erfüllen~~.	+	Nachdem <math>t=\sin x</math>, muss ''x'' der Gleichungen <math>\sin x = \tfrac{1}{2}</math> oder <math>\sin x = -1\,</math> genügen.


	<math>\sin x = \frac{1}{2}</math>:		<math>\sin x = \frac{1}{2}</math>:

-	Diese Gleichung hat die Lösungen <math>x = \pi/6</math> und <math>x = \pi - \pi/6 = 5\pi/6</math> im Einheitskreis~~, und daher~~ die allgemeinen Lösungen	+	Diese Gleichung hat die Lösungen <math>x = \pi/6</math> und <math>x = \pi - \pi/6 = 5\pi/6</math> auf dem Einheitskreis. Daher gilt für die allgemeinen Lösungen

-	{{Abgesetzte Formel\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{und}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x = \frac{\pi}{6}+2n\pi\qquad\text{und}\qquad x = \frac{5\pi}{6}+2n\pi\,.</math>}}


	<math>\sin x = -1</math>:		<math>\sin x = -1</math>:

-	~~Hat~~ nur die Lösung <math>x = 3\pi/2</math> im Einheitskreis, ~~und~~ daher die allgemeinen Lösungen	+	hat nur die Lösung <math>x = 3\pi/2</math> auf dem Einheitskreis, daher sind die allgemeinen Lösungen

-	{{Abgesetzte Formel\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x = \frac{3\pi}{2} + 2n\pi\,.</math>}}