Solution 4.4:6b - Revision history

Tek at 14:38, 13 October 2008

2008-10-13T14:38:27Z

←Older revision		Revision as of 14:38, 13 October 2008
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	After moving the terms over to the left-hand side, so that		After moving the terms over to the left-hand side, so that

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

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

		+	{{Displayed math\|\|<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}

-	~~we see~~ that ~~we can take out a common factor~~	+	and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
-	<math>\~~text{~~cos }x</math>,	+


-	<math>\cos x~~\left( \sqrt{2}\sin x-1 \right)~~=0</math>	+	<math>\cos x=0:</math>

		+	This basic equation has solutions <math>x=\pi/2</math> and <math>x=3\pi/2</math> in the unit circle, and from this we see that the general solution is

-	~~and that the equation is only satisfied if at least one of the factors,~~	+	{{Displayed math\|\|<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
-	~~<math>\text~~{~~cos }x</~~math>	+
-	or	+
-	<math>\~~sqrt~~{2}\text{~~sin~~ }x-\~~text~~{1}</math>	+
-	~~is zero. Thus, there are two cases:~~	+

		+	where ''n'' is an arbitrary integer. Because the angles <math>\pi/2</math> and
		+	<math>3\pi/2</math> differ by <math>\pi</math>, the solutions can be summarized as

-	~~<math>\text~~{~~cos }x=0</~~math~~>: This basic equation has solutions~~	+	{{Displayed math\|\|<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
-	<math>x={\pi }/{2}\;</math>	+
-	~~and~~	+
-	~~<math>x=3{\pi~~ }~~/{2~~}~~\;</math>~~	+
-	~~in the unit circle, and from this we see that the general solution is~~	+

		+	where ''n'' is an arbitrary integer.

-	<math>x=\frac{\pi }{2}+2n\pi </math>
-	and
-	<math>x=\frac{3\pi }{2}+2n\pi </math>

		+	<math>\sqrt{2}\sin x - 1 = 0:</math>

-	~~where~~	+	If we rearrange the equation, we obtain the basic equation as <math>\sin x = 1/\!\sqrt{2}</math>, which has the solutions <math>x=\pi/4</math> and <math>x=3\pi /4</math> in the unit circle and hence the general solution
-	<math>n\~~text~~{ }</math>	+
-	~~is an arbitrary integer. Because~~ the ~~angles~~	+
-	<math>{\pi }/~~{2}\;~~</math>	+
-	and	+
-	<math>3{\pi }/~~{2}\;~~</math>	+
-	~~differ by~~	+
-	~~<math>\pi </math>,~~ the ~~solutions can be summarized as~~	+

-		+	{{Displayed math\|\|<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
-	<math>x=\frac{\pi }{2}+n\pi </math>	+
-	(	+
-	~~<math>n</math>~~	+
-	~~an arbitrary integer).~~	+

-	√	+	where ''n'' is an arbitrary integer.
-	~~<math>\sqrt{2}\text{sin }x-\text{1}=0</math>~~	+
-	~~: if we rearrange the equation, we obtain the basic equation as~~	+
-	~~<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the solutions~~	+
-	~~<math>x={\pi }/{4}\;</math>~~	+
-	~~and~~	+
-	~~<math>x=3{\pi }/{4}\;</math>~~	+
-	~~in the unit circle and hence the general solution~~	+

-
-	<math>x=\frac{\pi }{4}+2n\pi </math>
-	and
-	<math>x=\frac{3\pi }{4}+2n\pi </math>
-
-
-	where
-	<math>n\text{ }</math>
-	can arbitrary integer.

	All in all, the original equation has the solutions		All in all, the original equation has the solutions

		+	{{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\text{ }</math>~~	+
-	an arbitrary integer).	+

Ian at 11:41, 1 October 2008

2008-10-01T11:41:33Z

←Older revision		Revision as of 11:41, 1 October 2008
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-	{{~~NAVCONTENT_START}~~}	+	After moving the terms over to the left-hand side, so that
-	<~~center~~> ~~[[Image:4_4_6b-1~~(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\sqrt{2}\sin x\cos x-\cos x=0</math>
-	<~~center~~> ~~[[Image:4_4_6b-~~2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	we see that we can take out a common factor
		+	<math>\text{cos }x</math>,
		+
		+
		+	<math>\cos x\left( \sqrt{2}\sin x-1 \right)=0</math>
		+
		+
		+	and that the equation is only satisfied if at least one of the factors,
		+	<math>\text{cos }x</math>
		+	or
		+	<math>\sqrt{2}\text{sin }x-\text{1}</math>
		+	is zero. Thus, there are two cases:
		+
		+
		+	<math>\text{cos }x=0</math>: This basic equation has solutions
		+	<math>x={\pi }/{2}\;</math>
		+	and
		+	<math>x=3{\pi }/{2}\;</math>
		+	in the unit circle, and from this we see that the general solution is
		+
		+
		+	<math>x=\frac{\pi }{2}+2n\pi </math>
		+	and
		+	<math>x=\frac{3\pi }{2}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer. Because the angles
		+	<math>{\pi }/{2}\;</math>
		+	and
		+	<math>3{\pi }/{2}\;</math>
		+	differ by
		+	<math>\pi </math>, the solutions can be summarized as
		+
		+
		+	<math>x=\frac{\pi }{2}+n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).
		+
		+	√
		+	<math>\sqrt{2}\text{sin }x-\text{1}=0</math>
		+	: if we rearrange the equation, we obtain the basic equation as
		+	<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the solutions
		+	<math>x={\pi }/{4}\;</math>
		+	and
		+	<math>x=3{\pi }/{4}\;</math>
		+	in the unit circle and hence the general solution
		+
		+
		+	<math>x=\frac{\pi }{4}+2n\pi </math>
		+	and
		+	<math>x=\frac{3\pi }{4}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	can arbitrary integer.
		+
		+	All in all, the original equation has the solutions
		+
		+
		+	<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\text{ }</math>
		+	an arbitrary integer).

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

2008-09-09T10:07:03Z

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

←Older revision

Revision as of 10:07, 9 September 2008

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

2008-08-21T07:20:39Z

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

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2008-04-03T12:39:44Z

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