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Lösung 4.4:6b

Aus Online Mathematik Brückenkurs 1

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K (Lösning 4.4:6b moved to Solution 4.4:6b: Robot: moved page)
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After moving the terms over to the left-hand side, so that
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<center> [[Image:4_4_6b-1(2).gif]] </center>
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<math>\sqrt{2}\sin x\cos x-\cos x=0</math>
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<center> [[Image:4_4_6b-2(2).gif]] </center>
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we see that we can take out a common factor
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<math>\text{cos }x</math>,
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<math>\cos x\left( \sqrt{2}\sin x-1 \right)=0</math>
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and that the equation is only satisfied if at least one of the factors,
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<math>\text{cos }x</math>
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or
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<math>\sqrt{2}\text{sin }x-\text{1}</math>
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is zero. Thus, there are two cases:
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<math>\text{cos }x=0</math>: This basic equation has solutions
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<math>x={\pi }/{2}\;</math>
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and
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<math>x=3{\pi }/{2}\;</math>
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in the unit circle, and from this we see that the general solution is
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<math>x=\frac{\pi }{2}+2n\pi </math>
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and
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<math>x=\frac{3\pi }{2}+2n\pi </math>
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where
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<math>n\text{ }</math>
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is an arbitrary integer. Because the angles
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<math>{\pi }/{2}\;</math>
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and
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<math>3{\pi }/{2}\;</math>
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differ by
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<math>\pi </math>, the solutions can be summarized as
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<math>x=\frac{\pi }{2}+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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√
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<math>\sqrt{2}\text{sin }x-\text{1}=0</math>
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: if we rearrange the equation, we obtain the basic equation as
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<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the solutions
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<math>x={\pi }/{4}\;</math>
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and
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<math>x=3{\pi }/{4}\;</math>
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in the unit circle and hence the general solution
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<math>x=\frac{\pi }{4}+2n\pi </math>
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and
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<math>x=\frac{3\pi }{4}+2n\pi </math>
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 +
 
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where
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<math>n\text{ }</math>
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can arbitrary integer.
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All in all, the original equation has the solutions
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<math>\left\{ \begin{array}{*{35}l}
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x=\frac{\pi }{4}+2n\pi \\
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x=\frac{\pi }{2}+n\pi \\
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x=\frac{3\pi }{4}+2n\pi \\
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\end{array} \right.</math>
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(
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<math>n\text{ }</math>
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an arbitrary integer).

Version vom 11:41, 1. Okt. 2008

After moving the terms over to the left-hand side, so that


2sinxcosx−cosx=0 


we see that we can take out a common factor cos x,


cosx2sinx−1=0 


and that the equation is only satisfied if at least one of the factors, cos x or 2sin x−1  is zero. Thus, there are two cases:


cos x=0: This basic equation has solutions x=2 and x=32 in the unit circle, and from this we see that the general solution is


x=2+2n and x=23+2n


where n is an arbitrary integer. Because the angles 2 and 32 differ by , the solutions can be summarized as


x=2+n ( n an arbitrary integer).

√ 2sin x−1=0 

if we rearrange the equation, we obtain the basic equation as

sin x =12 , which has the solutions x=4 and x=34 in the unit circle and hence the general solution


x=4+2n and x=43+2n


where n can arbitrary integer.

All in all, the original equation has the solutions


x=4+2nx=2+nx=43+2n ( n an arbitrary integer).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:6b“