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Solution 4.4:2a

From Förberedande kurs i matematik 1

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m (Lösning 4.4:2a moved to Solution 4.4:2a: Robot: moved page)
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{{NAVCONTENT_START}}
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We draw a unit circle and mark on those angles on the circle which have a
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<center> [[Image:4_4_2a-1(2).gif]] </center>
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<math>y</math>
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{{NAVCONTENT_STOP}}
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-coordinate of
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{{NAVCONTENT_START}}
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<math>{\sqrt{3}}/{2}\;</math>, in order to see which solutions lie between
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<center> [[Image:4_4_2a-2(2).gif]] </center>
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<math>0</math>
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{{NAVCONTENT_STOP}}
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and
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<math>2\pi </math>.
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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
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<math>x={\pi }/{3}\;</math>
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as the angle which has a sine value of
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<math>{\sqrt{3}}/{2}\;</math>
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and then we have the reflectionally symmetric solution
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<math>x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}</math>
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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
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<math>2\pi </math>
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<math>x=\frac{\pi }{3}+2n\pi </math>
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and
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<math>x=\frac{2\pi }{3}+2n\pi </math>
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where
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<math>n</math>
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is an arbitrary integer.
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NOTE: when we write that the complete solution is given by
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<math>x=\frac{\pi }{3}+2n\pi </math>
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and
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<math>x=\frac{2\pi }{3}+2n\pi </math>,
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this means that for every integer
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<math>n</math>, we obtain a solution to the equation:
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<math>\begin{array}{*{35}l}
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n=0 & x=\frac{\pi }{3} & x=\frac{2\pi }{3} \\
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n=-1 & x=\frac{\pi }{3}+\left( -1 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -1 \right)\centerdot 2\pi \\
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n=1 & x=\frac{\pi }{3}+1\centerdot 2\pi & x=\frac{2\pi }{3}+1\centerdot 2\pi \\
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n=-2 & x=\frac{\pi }{3}+\left( -2 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -2 \right)\centerdot 2\pi \\
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n=2 & x=\frac{\pi }{3}+2\centerdot 2\pi & x=\frac{2\pi }{3}+2\centerdot 2\pi \\
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\end{array}</math>
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and so on.

Revision as of 13:30, 30 September 2008

We draw a unit circle and mark on 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=32 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+2n and x=32+2n


where n is an arbitrary integer.

NOTE: when we write that the complete solution is given by


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

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


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

and so on.

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