From Förberedande kurs i matematik 1

Revision as of 12:40, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_7b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_7b-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:44, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we use the Pythagorean identity and write <math>\sin^2\!x</math> as <math>1-\cos^2\!x</math>, the whole equation can be written in terms of <math>\cos x</math>,
-	<center> [[Bild:4_4_7b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>2(1-\cos^2\!x) - 3\cos x = 0\,,</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_4_7b-2(2).gif]] </center>	+	or, in rearranged form,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}</math>}}
		+
		+	With the equation expressed entirely in terms of <math>\cos x</math>, we can introduce a new unknown variable <math>t=\cos x</math> and solve the equation with respect to ''t''. Expressed in terms of ''t'', the equation is
		+
		+	{{Displayed math||<math>2t^2+3t-2 = 0</math>}}
		+
		+	and this quadratic equation has the solutions <math>t=\tfrac{1}{2}</math> and
		+	<math>t=-2\,</math>.
		+
		+	In terms of ''x'', this means that either <math>\cos x = \tfrac{1}{2}</math> or <math>\cos x = -2</math>. The first case occurs when
		+
		+	{{Displayed math||<math>x=\pm \frac{\pi}{3}+2n\pi\qquad</math>(''n'' is an arbitrary integer),}}
		+
		+	whilst the equation <math>\cos x = -2</math> has no solutions at all (the values of cosine lie between -1 and 1).
		+
		+	The answer is that the equation has the solutions
		+
		+	{{Displayed math||<math>x = \pm\frac{\pi}{3} + 2n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer.

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Current revision

If we use the Pythagorean identity and write \displaystyle \sin^2\!x as \displaystyle 1-\cos^2\!x, the whole equation can be written in terms of \displaystyle \cos x,

\displaystyle 2(1-\cos^2\!x) - 3\cos x = 0\,,

or, in rearranged form,

\displaystyle 2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}

With the equation expressed entirely in terms of \displaystyle \cos x, we can introduce a new unknown variable \displaystyle t=\cos x and solve the equation with respect to t. Expressed in terms of t, the equation is

\displaystyle 2t^2+3t-2 = 0

and this quadratic equation has the solutions \displaystyle t=\tfrac{1}{2} and \displaystyle t=-2\,.

In terms of x, this means that either \displaystyle \cos x = \tfrac{1}{2} or \displaystyle \cos x = -2. The first case occurs when

\displaystyle x=\pm \frac{\pi}{3}+2n\pi\qquad(n is an arbitrary integer),

whilst the equation \displaystyle \cos x = -2 has no solutions at all (the values of cosine lie between -1 and 1).

The answer is that the equation has the solutions

\displaystyle x = \pm\frac{\pi}{3} + 2n\pi\,,

where n is an arbitrary integer.