From Förberedande kurs i matematik 1

Revision as of 12:47, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_8b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:15, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Suppose that <math>\cos x\ne 0</math>, so that we can divide both sides by <math>\cos x</math> to obtain
-	<center> [[Bild:4_4_8b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\frac{\sin x}{\cos x} = \sqrt{3}\qquad\text{i.e.}\qquad \tan x = \sqrt{3}\,\textrm{.}</math>}}
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		+	This equation has the solutions <math>x = \pi/3+n\pi</math> for all integers ''n''.
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		+	If, on the other hand, <math>\cos x=0</math>, then <math>\sin x = \pm 1</math> (draw a unit circle) and the equation cannot have such a solution.
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		+	Thus, the equation has the solutions
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		+	{{Displayed math||<math>x = \frac{\pi}{3}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}

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Current revision

Suppose that \displaystyle \cos x\ne 0, so that we can divide both sides by \displaystyle \cos x to obtain

\displaystyle \frac{\sin x}{\cos x} = \sqrt{3}\qquad\text{i.e.}\qquad \tan x = \sqrt{3}\,\textrm{.}

This equation has the solutions \displaystyle x = \pi/3+n\pi for all integers n.

If, on the other hand, \displaystyle \cos x=0, then \displaystyle \sin x = \pm 1 (draw a unit circle) and the equation cannot have such a solution.

Thus, the equation has the solutions

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