Solution 4.4:2f - Revision history

Tek at 12:44, 13 October 2008

Tek — Mon, 13 Oct 2008 12:44:31 GMT

←Older revision		Revision as of 12:44, 13 October 2008
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-	Using the unit circle shows that the equation	+	Using the unit circle shows that the equation <math>\cos 3x = -1/\!\sqrt{2}</math>
-	<math>\~~text{~~cos ~~3}x~~=-~~\frac{~~1}{\sqrt{2}}</math>	+	has two solutions for <math>0\le 3x\le 2\pi\,</math>,
-	has two solutions for	+
-	<math>0\le ~~\text{3}x~~\le ~~\text{~~2}\pi </math>,	+

-		+	{{Displayed math\|\|<math>3x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\qquad\text{and}\qquad 3x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\,\textrm{.}</math>}}
-	<math>3x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}~~</math>~~	+
-	and	+
-	~~<math>~~3x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}</math>	+

	[[Image:4_4_2_f.gif\|center]]		[[Image:4_4_2_f.gif\|center]]

-	We obtain the other solutions by adding multiples of	+	We obtain the other solutions by adding multiples of <math>2\pi</math>,
-	<math>2\pi </math>,	+
-		+
-		+
-	~~<math>3x=\frac{3\pi }{4}+2n\pi </math>~~	+
-	~~and~~	+
-	~~<math>3x=\frac{5\pi }{4}+2n\pi </math>~~	+

		+	{{Displayed math\|\|<math>3x = \frac{3\pi}{4} + 2n\pi\qquad\text{and}\qquad 3x = \frac{5\pi}{4} + 2n\pi\,,</math>}}

	i.e.		i.e.

		+	{{Displayed math\|\|<math>x = \frac{\pi}{4} + \frac{2}{3}n\pi\qquad\text{and}\qquad x = \frac{5\pi}{12} + \frac{2}{3}n\pi\,,</math>}}

-	~~<math>x=\frac{\pi }{4}+\frac{2}{3}n\pi </math>~~	+	where ''n'' is an arbitrary integer.
-	~~and~~	+
-	~~<math>x=\frac{5\pi }{12}+\frac{2}{3}n\pi </math>~~	+
-		+
-		+
-	where	+
-	~~<math>~~n~~</math>~~	+
-	is an arbitrary integer.	+

Ian — Wed, 01 Oct 2008 08:57:28 GMT

←Older revision		Revision as of 08:57, 1 October 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	Using the unit circle shows that the equation
-	<~~center~~> ~~[[Image:4_4_2f.gif]]~~ </~~center~~>	+	<math>\text{cos 3}x=-\frac{1}{\sqrt{2}}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	has two solutions for
		+	<math>0\le \text{3}x\le \text{2}\pi </math>,
		+
		+
		+	<math>3x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}</math>
		+	and
		+	<math>3x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}</math>

	[[Image:4_4_2_f.gif\|center]]		[[Image:4_4_2_f.gif\|center]]
		+
		+	We obtain the other solutions by adding multiples of
		+	<math>2\pi </math>,
		+
		+
		+	<math>3x=\frac{3\pi }{4}+2n\pi </math>
		+	and
		+	<math>3x=\frac{5\pi }{4}+2n\pi </math>
		+
		+
		+	i.e.
		+
		+
		+	<math>x=\frac{\pi }{4}+\frac{2}{3}n\pi </math>
		+	and
		+	<math>x=\frac{5\pi }{12}+\frac{2}{3}n\pi </math>
		+
		+
		+	where
		+	<math>n</math>
		+	is an arbitrary integer.

Tekbot — Tue, 09 Sep 2008 10:03:43 GMT

←Older revision

Revision as of 10:03, 9 September 2008

Tekbot — Thu, 21 Aug 2008 07:18:59 GMT

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

Madde — Tue, 19 Aug 2008 12:17:54 GMT

Oskar — Thu, 03 Apr 2008 12:29:41 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_2f.gif </center> {{NAVCONTENT_STOP}}

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