Lösung 4.4:8b

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Suppose that
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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
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<math>\text{cos }x\ne 0</math>
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, so that we can divide both sides by
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<math>\text{cos }x</math>
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to obtain
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{{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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<math>\frac{\sin x}{\cos x}=\sqrt{3}</math>
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i.e.
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<math>\tan x=\sqrt{3}</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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This equation has the solutions
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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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<math>x=\frac{\pi }{3}+n\pi </math>
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for all integers
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<math>n</math>.
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If, on the other hand,
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<math>\text{cos }x=0</math>, so
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<math>\text{sin }x\text{ }=\pm \text{1}</math>
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( draw a unit circle) and the equation cannot have such a solution.
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Thus, the equation has the solutions
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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<math>x=\frac{\pi }{3}+n\pi </math>
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(
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<math>n</math>
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an arbitrary integer).
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Version vom 08:15, 14. Okt. 2008

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

Vorlage:Displayed math

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

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:8b“