Solution 4.4:2d

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m (Lösning 4.4:2d moved to Solution 4.4:2d: Robot: moved page)
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Apart from the fact that there is a
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<center> [[Image:4_4_2d-1(2).gif]] </center>
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<math>\text{5}x</math>, this is a normal trigonometric equation of the type
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<math>\text{sin }y\text{ }=a</math> . If we are only interested in solutions which satisfy
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<math>0\le \text{5}x\le \text{2}\pi </math>, then a sketch of the unit circle shows that there are two such solutions,
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<center> [[Image:4_4_2d-2(2).gif]] </center>
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<math>\text{5}x=\text{ }\frac{\pi }{4}</math>
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and the reflectionally symmetric solution
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<math>\text{5}x=\text{ }\pi \text{-}\frac{\pi }{4}=\frac{3\pi }{4}</math>.
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[[Image:4_4_2_d.gif|center]]
[[Image:4_4_2_d.gif|center]]
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All of the equation's solutions are obtained from all values of 5x which differ by a multiple of 2π from either of these two solutions:
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<math>\text{5}x=\text{ }\frac{\pi }{4}+2n\pi </math>
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and
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<math>\text{5}x=\frac{3\pi }{4}+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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If we divide both of these by
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<math>\text{5}</math>, we obtain the solutions expressed in terms of x alone:
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<math>x=\frac{\pi }{20}+\frac{2}{5}n\pi </math>
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and
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<math>x=\frac{3\pi }{20}+\frac{2}{5}n\pi </math>,
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where
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<math>n</math>
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is an arbitrary integer.

Revision as of 13:55, 30 September 2008

Apart from the fact that there is a \displaystyle \text{5}x, this is a normal trigonometric equation of the type \displaystyle \text{sin }y\text{ }=a . If we are only interested in solutions which satisfy \displaystyle 0\le \text{5}x\le \text{2}\pi , then a sketch of the unit circle shows that there are two such solutions, \displaystyle \text{5}x=\text{ }\frac{\pi }{4} and the reflectionally symmetric solution \displaystyle \text{5}x=\text{ }\pi \text{-}\frac{\pi }{4}=\frac{3\pi }{4}.


All of the equation's solutions are obtained from all values of 5x which differ by a multiple of 2π from either of these two solutions:


\displaystyle \text{5}x=\text{ }\frac{\pi }{4}+2n\pi and \displaystyle \text{5}x=\frac{3\pi }{4}+2n\pi ,

where \displaystyle n is an arbitrary integer.

If we divide both of these by \displaystyle \text{5}, we obtain the solutions expressed in terms of x alone:


\displaystyle x=\frac{\pi }{20}+\frac{2}{5}n\pi and \displaystyle x=\frac{3\pi }{20}+\frac{2}{5}n\pi ,

where \displaystyle n is an arbitrary integer.

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