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Solution 4.4:6a

From Förberedande kurs i matematik 1

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m (Lösning 4.4:6a moved to Solution 4.4:6a: Robot: moved page)
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If we move everything over to the left-hand side,
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<center> [[Image:4_4_6a.gif]] </center>
+
 
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{{NAVCONTENT_STOP}}
+
 
 +
<math>\sin x\cos 3x-2\sin x=0</math>
 +
 
 +
 
 +
we see that both terms have
 +
<math>\text{sin }x\text{ }</math>
 +
as a common factor which we can take out:
 +
 
 +
 
 +
<math>\text{sin }x\text{ }\left( \cos 3x-2 \right)=0</math>
 +
 
 +
 
 +
In this factorized version of the equation, we see the equation has a solution only when one of the factors
 +
<math>\text{sin }x</math>
 +
or
 +
<math>\cos 3x-2</math>
 +
is zero. The factor
 +
<math>\text{sin }x</math>
 +
is zero for all values of
 +
<math>x</math>
 +
that are given by
 +
 
 +
 
 +
<math>x=n\pi </math>
 +
(
 +
<math>n</math>
 +
an arbitrary integer)
 +
 
 +
(see exercise 3.5:2c). The other factor
 +
<math>\cos 3x-2</math>
 +
can never be zero because the value of a cosine always lies between
 +
<math>-\text{1 }</math>
 +
and
 +
<math>\text{1}</math>, which gives that the largest value of
 +
<math>\cos 3x-2</math>
 +
is
 +
<math>-\text{1 }</math>.
 +
 
 +
The solutions are therefore
 +
 
 +
 +
<math>x=n\pi </math>
 +
(
 +
<math>n</math>
 +
an arbitrary integer).

Revision as of 11:25, 1 October 2008

If we move everything over to the left-hand side,


sinxcos3x−2sinx=0


we see that both terms have sin x as a common factor which we can take out:


sin x cos3x−2=0 


In this factorized version of the equation, we see the equation has a solution only when one of the factors sin x or cos3x−2 is zero. The factor sin x is zero for all values of x that are given by


x=n ( n an arbitrary integer)

(see exercise 3.5:2c). The other factor cos3x−2 can never be zero because the value of a cosine always lies between −1 and 1, which gives that the largest value of cos3x−2 is −1 .

The solutions are therefore


x=n ( n an arbitrary integer).

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