Aus Online Mathematik Brückenkurs 1

Version vom 08:15, 14. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 09:01, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
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	Suppose that <math>\cos x\ne 0</math>, so that we can divide both sides by <math>\cos x</math> to obtain		Suppose that <math>\cos x\ne 0</math>, so that we can divide both sides by <math>\cos x</math> to obtain
-	{{Displayed math||<math>\frac{\sin x}{\cos x} = \sqrt{3}\qquad\text{i.e.}\qquad \tan x = \sqrt{3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>\frac{\sin x}{\cos x} = \sqrt{3}\qquad\text{i.e.}\qquad \tan x = \sqrt{3}\,\textrm{.}</math>}}
	This equation has the solutions <math>x = \pi/3+n\pi</math> for all integers ''n''.		This equation has the solutions <math>x = \pi/3+n\pi</math> for all integers ''n''.
Zeile 9:		Zeile 9:
	Thus, the equation has the solutions		Thus, the equation has the solutions
-	{{Displayed math||<math>x = \frac{\pi}{3}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}	+	{{Abgesetzte Formel||<math>x = \frac{\pi}{3}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}

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Version vom 09:01, 22. Okt. 2008

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).