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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
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<center> [[Image:4_4_2a-1(2).gif]] </center>
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<math>0</math> and <math>2\pi</math>.
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<center> [[Image:4_4_2a-2(2).gif]] </center>
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[[Image:4_4_2_a.gif|center]]
[[Image:4_4_2_a.gif|center]]
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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}
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&n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt]
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&n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt]
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&n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt]
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&n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt]
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&n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt]
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&\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots
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\end{array}</math>}}

Current revision

We draw a unit circle and mark those angles on the circle which have a y-coordinate of 32 , in order to see which solutions lie between 0 and 2.

In the first quadrant, we recognize x=3 as the angle which has a sine value of 32  and then we have the reflectionally symmetric solution x=−3=23 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 2

x=3+2nandx=32+2n

where n is an arbitrary integer.


Note: When we write that the complete solution is given by

x=3+2nandx=32+2n,

this means that for every integer n, we obtain a solution to the equation:

n=0:n=−1:n=1:n=−2:n=2:x=3x=3+(−1)2x=3+12x=3+(−2)2x=3+22x=32x=32+(−1)2x=32+12x=32+(−2)2x=32+22
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