Lösung 4.4:8b

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Suppose that
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<center> [[Image:4_4_8b.gif]] </center>
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<math>\text{cos }x\ne 0</math>
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{{NAVCONTENT_STOP}}
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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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<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
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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
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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).

Version vom 13:39, 1. Okt. 2008

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


\displaystyle \frac{\sin x}{\cos x}=\sqrt{3} i.e. \displaystyle \tan x=\sqrt{3}


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

If, on the other hand, \displaystyle \text{cos }x=0, so \displaystyle \text{sin }x\text{ }=\pm \text{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 ( \displaystyle n an arbitrary integer).

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