From Förberedande kurs i matematik 1

Revision as of 12:27, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_2d-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_2d-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:32, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Apart from the fact that there is a <math>5x</math>, this is a normal trigonometric equation of the type <math>\sin y = a\,</math>. If we are only interested in solutions which satisfy <math>0\le 5x\le 2\pi</math>, then a sketch of the unit circle shows that there are two such solutions, <math>5x = \pi/4</math> and the reflectionally symmetric solution <math>5x = \pi - \pi/4 = 3\pi/4\,</math>.
-	<center> [[Bild:4_4_2d-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	[[Image:4_4_2_d.gif|center]]
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_4_2d-2(2).gif]] </center>	+	All of the equation's solutions are obtained from all values of <math>5x</math> which differ by a multiple of <math>2\pi</math> from either of these two solutions,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>5x = \frac{\pi}{4} + 2n\pi\qquad\text{and}\qquad 5x = \frac{3\pi}{4} + 2n\pi\,,</math>}}
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		+	where ''n'' is an arbitrary integer.
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		+	If we divide both of these by 5, we obtain the solutions expressed in terms of ''x'' alone,
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		+	{{Displayed math||<math>x = \frac{\pi}{20} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{3\pi}{20} + \frac{2}{5}n\pi\,,</math>}}
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		+	where ''n'' is an arbitrary integer.

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Current revision

Apart from the fact that there is a \displaystyle 5x, this is a normal trigonometric equation of the type \displaystyle \sin y = a\,. If we are only interested in solutions which satisfy \displaystyle 0\le 5x\le 2\pi, then a sketch of the unit circle shows that there are two such solutions, \displaystyle 5x = \pi/4 and the reflectionally symmetric solution \displaystyle 5x = \pi - \pi/4 = 3\pi/4\,.

All of the equation's solutions are obtained from all values of \displaystyle 5x which differ by a multiple of \displaystyle 2\pi from either of these two solutions,

\displaystyle 5x = \frac{\pi}{4} + 2n\pi\qquad\text{and}\qquad 5x = \frac{3\pi}{4} + 2n\pi\,,

where n is an arbitrary integer.

If we divide both of these by 5, we obtain the solutions expressed in terms of x alone,

\displaystyle x = \frac{\pi}{20} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{3\pi}{20} + \frac{2}{5}n\pi\,,

where n is an arbitrary integer.