Solution 4.4:6a

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Current revision (07:33, 21 January 2009) (edit) (undo)
(exercise 3.5:2c --> exercise 4.4:2c)
 
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If we move everything over to the left-hand side,
If we move everything over to the left-hand side,
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{{Displayed math||<math>\sin x\cos 3x-2\sin x=0</math>}}
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<math>\sin x\cos 3x-2\sin x=0</math>
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we see that both terms have <math>\sin x</math> as a common factor which we can take out,
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{{Displayed math||<math>\sin x (\cos 3x-2) = 0\,\textrm{.}</math>}}
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we see that both terms have
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In this factorized version of the equation, we see the equation has a solution only when one of the factors <math>\sin x</math> or <math>\cos 3x-2</math> is zero. The factor <math>\sin x</math> is zero for all values of ''x'' that are given by
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<math>\text{sin }x\text{ }</math>
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as a common factor which we can take out:
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{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
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<math>\text{sin }x\text{ }\left( \cos 3x-2 \right)=0</math>
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(see exercise 4.4:2c). The other factor <math>\cos 3x-2</math> can never be zero because the value of a cosine always lies between <math>-1</math> and <math>1</math>, which gives that the largest value of <math>\cos 3x-2</math> is <math>-1</math>.
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In this factorized version of the equation, we see the equation has a solution only when one of the factors
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<math>\text{sin }x</math>
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or
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<math>\cos 3x-2</math>
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is zero. The factor
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<math>\text{sin }x</math>
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is zero for all values of
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<math>x</math>
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that are given by
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<math>x=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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(see exercise 3.5:2c). The other factor
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<math>\cos 3x-2</math>
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can never be zero because the value of a cosine always lies between
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<math>-\text{1 }</math>
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and
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<math>\text{1}</math>, which gives that the largest value of
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<math>\cos 3x-2</math>
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is
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<math>-\text{1 }</math>.
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The solutions are therefore
The solutions are therefore
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{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
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<math>x=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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Current revision

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

\displaystyle \sin x\cos 3x-2\sin x=0

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

\displaystyle \sin x (\cos 3x-2) = 0\,\textrm{.}

In this factorized version of the equation, we see the equation has a solution only when one of the factors \displaystyle \sin x or \displaystyle \cos 3x-2 is zero. The factor \displaystyle \sin x is zero for all values of x that are given by

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer)}

(see exercise 4.4:2c). The other factor \displaystyle \cos 3x-2 can never be zero because the value of a cosine always lies between \displaystyle -1 and \displaystyle 1, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -1.

The solutions are therefore

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer).}
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