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Solution 4.4:6a

From Förberedande kurs i matematik 1

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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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{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer)}</math>}}
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(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>-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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(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>.
The solutions are therefore
The solutions are therefore
{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
{{Displayed math||<math>x=n\pi\qquad\text{(n is an arbitrary integer).}</math>}}

Current revision

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

sinxcos3x−2sinx=0

we see that both terms have sinx as a common factor which we can take out,

sinx(cos3x−2)=0.

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

x=n(n is an arbitrary integer)

(see exercise 4.4:2c). 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 is an arbitrary integer).
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