Solution 4.4:7a - Revision history

Tek at 07:32, 14 October 2008

2008-10-14T07:32:07Z

←Older revision		Revision as of 07:32, 14 October 2008
Line 1:		Line 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 at 12:49, 1 October 2008

2008-10-01T12:49:28Z

←Older revision		Revision as of 12:49, 1 October 2008
Line 1:		Line 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

←Older revision

Revision as of 10:07, 9 September 2008

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

2008-08-21T07:20:59Z

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

←Older revision		Revision as of 07:20, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_4_7a-1(2).gif]] </center>	+	<center> [[Image:4_4_7a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_4_7a-2(2).gif]] </center>	+	<center> [[Image:4_4_7a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_7a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_4_7a-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T12:41:44Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_7a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_7a-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_4_7a-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:4_4_7a-2(2).gif]] </center>
{{NAVCONTENT_STOP}}