Solution 4.4:2a
From Förberedande kurs i matematik 1
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- | { | + | We draw a unit circle and mark those angles on the circle which have a ''y''-coordinate of <math>\sqrt{3}/2</math>, in order to see which solutions lie between |
- | < | + | <math>0</math> and <math>2\pi</math>. |
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+ | [[Image:4_4_2_a.gif|center]] | ||
- | [[ | + | In the first quadrant, we recognize <math>x = \pi/3</math> as the angle which has a sine value of <math>\sqrt{3}/2</math> and then we have the reflectionally symmetric solution <math>x = \pi - \pi/3 = 2\pi/3</math> in the second quadrant. |
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+ | Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of <math>2\pi</math> | ||
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+ | {{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,,</math>}} | ||
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+ | where ''n'' is an arbitrary integer. | ||
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+ | Note: When we write that the complete solution is given by | ||
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+ | {{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,}</math>}} | ||
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+ | this means that for every integer ''n'', we obtain a solution to the equation: | ||
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+ | {{Displayed math||<math>\begin{array}{llll} | ||
+ | &n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt] | ||
+ | &n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt] | ||
+ | &n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt] | ||
+ | &n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt] | ||
+ | &n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt] | ||
+ | &\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots | ||
+ | \end{array}</math>}} |
Current revision
We draw a unit circle and mark those angles on the circle which have a y-coordinate of \displaystyle \sqrt{3}/2, in order to see which solutions lie between \displaystyle 0 and \displaystyle 2\pi.
In the first quadrant, we recognize \displaystyle x = \pi/3 as the angle which has a sine value of \displaystyle \sqrt{3}/2 and then we have the reflectionally symmetric solution \displaystyle x = \pi - \pi/3 = 2\pi/3 in the second quadrant.
Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of \displaystyle 2\pi
\displaystyle x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,, |
where n is an arbitrary integer.
Note: When we write that the complete solution is given by
\displaystyle x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,} |
this means that for every integer n, we obtain a solution to the equation:
\displaystyle \begin{array}{llll}
&n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt] &n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt] &n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt] &n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt] &n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt] &\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots \end{array} |