Solution 4.4:2e - Revision history

Tek at 14:37, 10 October 2008

2008-10-10T14:37:09Z

←Older revision		Revision as of 14:37, 10 October 2008
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-	This is almost the same equation as in exercise d. First, we determine the solutions to the equation	+	This is almost the same equation as in exercise d. First, we determine the solutions to the equation when <math>0\le 5x\le 2\pi</math>, and using the unit circle shows that there are two of these,
-	when	+
-	<math>0\le ~~\text{5}x~~\le ~~\text{~~2}\pi </math>, and using the unit circle shows that there are two of these:	+
-	~~<math>\text{5}x\text{ }=\frac{\pi }{6}</math>~~	+
-	~~and~~	+
-	~~<math>\text{5}x\text{ }=\pi -\frac{\pi }{6}=\frac{5\pi }{6}</math>.~~	+

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

	[[Image:4_4_2_e.gif\|center]]		[[Image:4_4_2_e.gif\|center]]

-	We obtain the remaining solutions by adding multiples of	+	We obtain the remaining solutions by adding multiples of <math>2\pi</math> to the two solutions above,
-	<math>2\pi </math>	+
-	to the two solutions above:	+

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

-	~~<math>\text{5}x\text{ }=\frac{\pi }{6}+2n\pi </math>~~	+	where ''n'' is an arbitrary integer, or if we divide by 5,
-	~~and~~	+
-	~~<math>\text{5}x\text{ }=\frac{5\pi }{6}+2n\pi </math>~~	+
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer),	+

-	~~or if we divide by~~	+	{{Displayed math\|\|<math>x = \frac{\pi}{30} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{\pi}{6} + \frac{2}{5}n\pi\,,</math>}}
-	<math>\text{5}</math>:	+

-		+	where ''n'' is an arbitrary integer.
-	~~<math>x\text{ }=\frac{\pi }{30}+\frac{2}{5}~~n~~\pi </math>~~	+
-	~~and~~	+
-	~~<math>x\text{ }=\frac{\pi }{6}+\frac{2}{5}n\pi </math>~~	+
-	(	+
-	~~<math>n</math>~~	+
-	an arbitrary integer).	+

Ian at 08:36, 1 October 2008

2008-10-01T08:36:21Z

←Older revision		Revision as of 08:36, 1 October 2008
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-	{{~~NAVCONTENT_START~~}}	+	This is almost the same equation as in exercise d. First, we determine the solutions to the equation
-	<~~center~~> ~~[[Image~~:~~4_4_2e.gif]]~~ </~~center~~>	+	when
-	{{~~NAVCONTENT_STOP~~}}	+	<math>0\le \text{5}x\le \text{2}\pi </math>, and using the unit circle shows that there are two of these:
		+	<math>\text{5}x\text{ }=\frac{\pi }{6}</math>
		+	and
		+	<math>\text{5}x\text{ }=\pi -\frac{\pi }{6}=\frac{5\pi }{6}</math>.
		+

	[[Image:4_4_2_e.gif\|center]]		[[Image:4_4_2_e.gif\|center]]
		+
		+	We obtain the remaining solutions by adding multiples of
		+	<math>2\pi </math>
		+	to the two solutions above:
		+
		+
		+	<math>\text{5}x\text{ }=\frac{\pi }{6}+2n\pi </math>
		+	and
		+	<math>\text{5}x\text{ }=\frac{5\pi }{6}+2n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer),
		+
		+	or if we divide by
		+	<math>\text{5}</math>:
		+
		+
		+	<math>x\text{ }=\frac{\pi }{30}+\frac{2}{5}n\pi </math>
		+	and
		+	<math>x\text{ }=\frac{\pi }{6}+\frac{2}{5}n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:2e moved to Solution 4.4:2e: Robot: moved page

2008-09-09T10:03:23Z

Lösning 4.4:2e moved to Solution 4.4:2e: Robot: moved page

←Older revision

Revision as of 10:03, 9 September 2008

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

2008-08-21T07:18:49Z

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

←Older revision		Revision as of 07:18, 21 August 2008
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-	[[~~Bild~~:4_4_2_e.gif\|center]]	+	[[Image:4_4_2_e.gif\|center]]

Madde at 12:17, 19 August 2008

2008-08-19T12:17:17Z

←Older revision		Revision as of 12:17, 19 August 2008
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		+	[[Bild:4_4_2_e.gif\|center]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_2e.gif {{NAVCONTENT_STOP}}

2008-04-03T12:29:05Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_2e.gif </center> {{NAVCONTENT_STOP}}

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