Solution 4.4:8a - Revision history

Tek at 08:12, 14 October 2008

2008-10-14T08:12:05Z

←Older revision		Revision as of 08:12, 14 October 2008
Line 1:		Line 1:
-	If we use the formula for double angles,	+	If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
-	<math>\~~text{~~sin ~~2}x~~=\~~text{2sin }~~x\~~text{~~ cos }x</math>, and move all the terms over to the left-hand side, the equation becomes	+

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

-	<math>~~2\sin x\cos x-\sqrt{2}~~\cos x~~=0.~~</math>	+	Then, we see that we can take a factor <math>\cos x</math> out of both terms,

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

-	~~Then, we see that we can take a factor~~ cos x ~~out of both terms~~,	+	and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.


-	<math>\cos x~~\left( 2\sin x-\sqrt{2} \right)~~=0</math>	+	<math>\cos x = 0</math>:

-		+	This equation has the general solution
-	~~and hence divide up the equation into two cases. The equation is satisfied either if~~	+
-	~~<math>\text{cos }x=0\text{ }</math>~~	+
-	~~or if~~	+
-	~~<math>2\sin x-\sqrt{2}=0</math>.~~	+
-		+
-		+
-	~~<math>\text{cos }x=0\text{ }</math>: this~~ equation has the general solution	+

		+	{{Displayed math\|\|<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}

-	<math>x=\frac{\pi }{2}+n\pi </math>
-	(
-	<math>n</math>
-	an arbitrary integer)

		+	<math>2\sin x-\sqrt{2}=0</math>:

-	~~<math>2\sin x-\sqrt{2}=0</math>:~~ If we collect	+	If we collect <math>\sin x</math> on the left-hand side, we obtain the equation
-	<math>\~~text{~~sin }x</math>	+	<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
-	on the left-hand side, we obtain the equation	+
-	<math>\~~text{~~sin }x~~\text{ }~~={1}/{\sqrt{2}~~}\;~~</math>, which has the general solution	+

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
		+	x &= \frac{3\pi}{4}+2n\pi\,,
		+	\end{align}\right.</math>}}

-	~~<math>\left\{ \begin{array}{*{35}l}~~	+	where ''n'' is an arbitrary integer.
-	~~x=\frac{\pi }{4}+2n\pi \\~~	+
-	~~x=\frac{3\pi }{4}+2n\pi \\~~	+
-	~~\end{array} \right.</math>~~	+
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer)	+

		+
	The complete solution of the equation is		The complete solution of the equation is

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
		+	x &= \frac{\pi}{2}+n\pi\,,\\[5pt]
		+	x &= \frac{3\pi}{4}+2n\pi\,,
		+	\end{align}\right.</math>}}

-	~~<math>\left\{ \begin{array}{*{35}l}~~	+	where ''n'' is an arbitrary integer.
-	~~x=\frac{\pi }{4}+2n\pi \\~~	+
-	~~x=\frac{\pi }{2}+~~n~~\pi \\~~	+
-	~~x=\frac{3\pi }{4}+2n\pi \\~~	+
-	~~\end{array} \right.</math>~~	+
-	(	+
-	~~<math>n</math>~~	+
-	an arbitrary integer).	+

Ian at 13:31, 1 October 2008

2008-10-01T13:31:42Z

←Older revision		Revision as of 13:31, 1 October 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	If we use the formula for double angles,
-	<~~center~~> ~~[[Image:4_4_8a~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\text{sin 2}x=\text{2sin }x\text{ cos }x</math>, and move all the terms over to the left-hand side, the equation becomes
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image~~:~~4_4_8a~~-2~~(2)~~.~~gif]]~~ </~~center~~>	+	<math>2\sin x\cos x-\sqrt{2}\cos x=0.</math>
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	Then, we see that we can take a factor cos x out of both terms,
		+
		+
		+	<math>\cos x\left( 2\sin x-\sqrt{2} \right)=0</math>
		+
		+
		+	and hence divide up the equation into two cases. The equation is satisfied either if
		+	<math>\text{cos }x=0\text{ }</math>
		+	or if
		+	<math>2\sin x-\sqrt{2}=0</math>.
		+
		+
		+	<math>\text{cos }x=0\text{ }</math>: this equation has the general solution
		+
		+
		+	<math>x=\frac{\pi }{2}+n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer)
		+
		+
		+	<math>2\sin x-\sqrt{2}=0</math>: If we collect
		+	<math>\text{sin }x</math>
		+	on the left-hand side, we obtain the equation
		+	<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the general solution
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{\pi }{4}+2n\pi \\
		+	x=\frac{3\pi }{4}+2n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer)
		+
		+	The complete solution of the equation is
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{\pi }{4}+2n\pi \\
		+	x=\frac{\pi }{2}+n\pi \\
		+	x=\frac{3\pi }{4}+2n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:8a moved to Solution 4.4:8a: Robot: moved page

2008-09-09T10:08:43Z

Lösning 4.4:8a moved to Solution 4.4:8a: Robot: moved page

←Older revision

Revision as of 10:08, 9 September 2008

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

2008-08-21T07:21:29Z

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

←Older revision		Revision as of 07:21, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_4_8a-1(2).gif]] </center>	+	<center> [[Image:4_4_8a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_4_8a-2(2).gif]] </center>	+	<center> [[Image:4_4_8a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_8a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_4_8a-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T12:42:53Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_8a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_8a-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_4_8a-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:4_4_8a-2(2).gif]] </center>
{{NAVCONTENT_STOP}}