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>}} |
+ | |||
+ | In exercise 4.4:5a, we saw that an equality of the type | ||
+ | |||
+ | {{Displayed math||<math>\sin u = \sin v</math>}} | ||
+ | |||
+ | is satisfied if | ||
+ | |||
+ | {{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 | ||
+ | |||
+ | {{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}} | ||
+ | |||
+ | i.e. | ||
+ | |||
+ | {{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}} | ||
+ | |||
+ | The solutions to the equation are thus | ||
+ | |||
+ | {{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 \displaystyle \sin (-x) = -\sin x, the equation can be rewritten as
\displaystyle \sin 2x = \sin (-x)\,\textrm{.} |
In exercise 4.4:5a, we saw that an equality of the type
\displaystyle \sin u = \sin v |
is satisfied if
\displaystyle u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,, |
where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
\displaystyle 2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,, |
i.e.
\displaystyle 3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.} |
The solutions to the equation are thus
\displaystyle \left\{\begin{align}
x &= \frac{2n\pi}{3}\,,\\[5pt] x &= \pi + 2n\pi\,, \end{align}\right. |
where n is an arbitrary integer.