Solution 4.4:6a
From Förberedande kurs i matematik 1
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(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>}} | ||
| - | (see exercise | + | (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,
| \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).} |
