Solution 4.4:8b - Revision history

Tek at 08:15, 14 October 2008

2008-10-14T08:15:34Z

←Older revision		Revision as of 08:15, 14 October 2008
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-	Suppose that	+	Suppose that <math>\cos x\ne 0</math>, so that we can divide both sides by <math>\cos x</math> to obtain
-	<math>\~~text{~~cos }x\ne 0</math>	+
-	, so that we can divide both sides by	+
-	<math>\~~text{~~cos }x</math>	+
-	to obtain	+

-		+	{{Displayed math\|\|<math>\frac{\sin x}{\cos x} = \sqrt{3}\qquad\text{i.e.}\qquad \tan x = \sqrt{3}\,\textrm{.}</math>}}
-	<math>\frac{\sin x}{\cos x}=\sqrt{3}~~</math>~~	+
-	i.e.	+
-	~~<math>~~\tan x=\sqrt{3}</math>	+

		+	This equation has the solutions <math>x = \pi/3+n\pi</math> for all integers ''n''.

-	~~This equation has the solutions~~	+	If, on the other hand, <math>\cos x=0</math>, then <math>\sin x = \pm 1</math> (draw a unit circle) and the equation cannot have such a solution.
-	~~<math>x=\frac{\pi }{3}+n\pi </math>~~	+
-	~~for all integers~~	+
-	~~<math>n</math>.~~	+
-		+
-	If, on the other hand,	+
-	<math>\~~text{~~cos }x=0</math>, so	+
-	<math>\~~text{~~sin }x~~\text{ }~~=\pm ~~\text{~~1}</math>	+
-	( draw a unit circle) and the equation cannot have such a solution.	+

	Thus, the equation has the solutions		Thus, the equation has the solutions

-		+	{{Displayed math\|\|<math>x = \frac{\pi}{3}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
-	<math>x=\frac{\pi }{3}+n\pi </math>	+
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer).	+

2008-10-01T13:39:09Z

←Older revision		Revision as of 13:39, 1 October 2008
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-	{{~~NAVCONTENT_START~~}}	+	Suppose that
-	<~~center~~> ~~[[Image:4_4_8b~~.~~gif]]~~ </~~center~~>	+	<math>\text{cos }x\ne 0</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, so that we can divide both sides by
		+	<math>\text{cos }x</math>
		+	to obtain
		+
		+
		+	<math>\frac{\sin x}{\cos x}=\sqrt{3}</math>
		+	i.e.
		+	<math>\tan x=\sqrt{3}</math>
		+
		+
		+	This equation has the solutions
		+	<math>x=\frac{\pi }{3}+n\pi </math>
		+	for all integers
		+	<math>n</math>.
		+
		+	If, on the other hand,
		+	<math>\text{cos }x=0</math>, so
		+	<math>\text{sin }x\text{ }=\pm \text{1}</math>
		+	( draw a unit circle) and the equation cannot have such a solution.
		+
		+	Thus, the equation has the solutions
		+
		+
		+	<math>x=\frac{\pi }{3}+n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

2008-09-09T10:09:03Z

←Older revision

Revision as of 10:09, 9 September 2008

2008-08-21T07:21:39Z

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

2008-04-03T12:47:41Z

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