From Förberedande kurs i matematik 1

Revision as of 07:20, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (07:33, 21 January 2009) (edit) (undo) Tek (Talk | contribs) (exercise 3.5:2c --> exercise 4.4:2c)
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-	{{NAVCONTENT_START}}	+	If we move everything over to the left-hand side,
-	<center> [[Image:4_4_6a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sin x\cos 3x-2\sin x=0</math>}}
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		+	we see that both terms have <math>\sin x</math> as a common factor which we can take out,
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		+	{{Displayed math||<math>\sin x (\cos 3x-2) = 0\,\textrm{.}</math>}}
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		+	In this factorized version of the equation, we see the equation has a solution only when one of the factors <math>\sin x</math> or <math>\cos 3x-2</math> is zero. The factor <math>\sin x</math> is zero for all values of ''x'' that are given by
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		+	{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
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		+	(see exercise 4.4:2c). The other factor <math>\cos 3x-2</math> can never be zero because the value of a cosine always lies between <math>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.
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		+	The solutions are therefore
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		+	{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}

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Current revision

If we move everything over to the left-hand side,

\displaystyle \sin x\cos 3x-2\sin x=0

we see that both terms have \displaystyle \sin x as a common factor which we can take out,

\displaystyle \sin x (\cos 3x-2) = 0\,\textrm{.}

In this factorized version of the equation, we see the equation has a solution only when one of the factors \displaystyle \sin x or \displaystyle \cos 3x-2 is zero. The factor \displaystyle \sin x is zero for all values of x that are given by

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer)}

(see exercise 4.4:2c). The other factor \displaystyle \cos 3x-2 can never be zero because the value of a cosine always lies between \displaystyle -1 and \displaystyle 1, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -1.

The solutions are therefore

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer).}