Solution 4.4:6a - Revision history

Tek: exercise 3.5:2c --> exercise 4.4:2c

Tek — Wed, 21 Jan 2009 07:33:16 GMT

exercise 3.5:2c --> exercise 4.4:2c

←Older revision		Revision as of 07:33, 21 January 2009
Line 11:		Line 11:
	{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}		{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}

-	(see exercise 3.5:2c). The other factor <math>\cos 3x-2</math> can never be zero because the value of a cosine always lies between <math>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.	+	(see exercise 4.4:2c). The other factor <math>\cos 3x-2</math> can never be zero because the value of a cosine always lies between <math>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.

	The solutions are therefore		The solutions are therefore

	{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}		{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}

Tek at 14:16, 13 October 2008

Tek — Mon, 13 Oct 2008 14:16:02 GMT

←Older revision		Revision as of 14:16, 13 October 2008
Line 1:		Line 1:
	If we move everything over to the left-hand side,		If we move everything over to the left-hand side,

		+	{{Displayed math\|\|<math>\sin x\cos 3x-2\sin x=0</math>}}

-	<math>\sin x~~\cos 3x-2\sin x=0~~</math>	+	we see that both terms have <math>\sin x</math> as a common factor which we can take out,

		+	{{Displayed math\|\|<math>\sin x (\cos 3x-2) = 0\,\textrm{.}</math>}}

-	we see ~~that both terms have~~	+	In this factorized version of the equation, we see the equation has a solution only when one of the factors <math>\sin x</math> or <math>\cos 3x-2</math> is zero. The factor <math>\sin x</math> is zero for all values of ''x'' that are given by
-	<math>\~~text{~~sin }x\~~text{ }~~</math>	+
-	~~as a common~~ factor ~~which we can take out:~~	+

		+	{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}

-	~~<math>\text{sin }x\text{ }\left( \cos 3x-2 \right)=0</math>~~	+	(see exercise 3.5:2c). The other factor <math>\cos 3x-2</math> can never be zero because the value of a cosine always lies between <math>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.
-		+
-		+
-	~~In this factorized version of the equation, we see the equation has a solution only when one of the factors~~	+
-	~~<math>\text{sin }x</math>~~	+
-	or	+
-	~~<math>\cos 3x-2</math>~~	+
-	~~is zero. The factor~~	+
-	~~<math>\text{sin }x</math>~~	+
-	~~is zero for all values of~~	+
-	~~<math>x</math>~~	+
-	~~that are given by~~	+
-		+
-		+
-	~~<math>x=n\pi </math>~~	+
-	(	+
-	~~<math>n</math>~~	+
-	~~an arbitrary integer)~~	+
-		+
-	(see exercise 3.5:2c). The other factor	+
-	<math>\cos 3x-2</math>	+
-	can never be zero because the value of a cosine always lies between	+
-	<math>-~~\text{~~1 }</math>	+
-	and	+
-	<math>~~\text{~~1}</math>, which gives that the largest value of	+
-	<math>\cos 3x-2</math>	+
-	is	+
-	<math>-~~\text{~~1 }</math>.	+

	The solutions are therefore		The solutions are therefore

-		+	{{Displayed math\|\|<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
-	<math>x=n\pi ~~</math>~~	+
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer).	+

Ian at 11:25, 1 October 2008

Ian — Wed, 01 Oct 2008 11:25:30 GMT

←Older revision		Revision as of 11:25, 1 October 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	If we move everything over to the left-hand side,
-	<~~center~~> ~~[[Image~~:~~4_4_6a~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\sin x\cos 3x-2\sin x=0</math>
		+
		+
		+	we see that both terms have
		+	<math>\text{sin }x\text{ }</math>
		+	as a common factor which we can take out:
		+
		+
		+	<math>\text{sin }x\text{ }\left( \cos 3x-2 \right)=0</math>
		+
		+
		+	In this factorized version of the equation, we see the equation has a solution only when one of the factors
		+	<math>\text{sin }x</math>
		+	or
		+	<math>\cos 3x-2</math>
		+	is zero. The factor
		+	<math>\text{sin }x</math>
		+	is zero for all values of
		+	<math>x</math>
		+	that are given by
		+
		+
		+	<math>x=n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer)
		+
		+	(see exercise 3.5:2c). The other factor
		+	<math>\cos 3x-2</math>
		+	can never be zero because the value of a cosine always lies between
		+	<math>-\text{1 }</math>
		+	and
		+	<math>\text{1}</math>, which gives that the largest value of
		+	<math>\cos 3x-2</math>
		+	is
		+	<math>-\text{1 }</math>.
		+
		+	The solutions are therefore
		+
		+
		+	<math>x=n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:6a moved to Solution 4.4:6a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:06:43 GMT

Lösning 4.4:6a moved to Solution 4.4:6a: Robot: moved page

←Older revision

Revision as of 10:06, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:20:29 GMT

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_6a.gif]] </center>	+	<center> [[Image:4_4_6a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_6a.gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 12:45:52 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_6a.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_4_6a.gif]] </center>
{{NAVCONTENT_STOP}}