Processing Math: Done
Solution 4.4:6c
From Förberedande kurs i matematik 1
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| - | {{ | + | If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\sin 2x = \sin (-x)\,\textrm{.}</math>}} |
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| + | In exercise 4.4:5a, we saw that an equality of the type | ||
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| + | {{Displayed math||<math>\sin u = \sin v</math>}} | ||
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| + | is satisfied if | ||
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| + | {{Displayed math||<math>u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,</math>}} | ||
| + | |||
| + | where ''n'' is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy | ||
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| + | {{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}} | ||
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| + | i.e. | ||
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| + | {{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}} | ||
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| + | The solutions to the equation are thus | ||
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| + | {{Displayed math||<math>\left\{\begin{align} | ||
| + | x &= \frac{2n\pi}{3}\,,\\[5pt] | ||
| + | x &= \pi + 2n\pi\,, | ||
| + | \end{align}\right.</math>}} | ||
| + | |||
| + | where ''n'' is an arbitrary integer. | ||
Current revision
If we use the trigonometric relation
In exercise 4.4:5a, we saw that an equality of the type
is satisfied if
 oru=−v+2n |
where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
| or2x=−(−x)+2n |
i.e.
| orx=+2n. |
The solutions to the equation are thus
   xx=32n=+2n |
where n is an arbitrary integer.
