Solution 4.4:3b

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Current revision (12:52, 13 October 2008) (edit) (undo)
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We see directly that
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We see directly that <math>x = \pi/5</math> is a solution to the equation, and using the unit circle we can also draw the conclusion that <math>x = \pi - \pi/5 = 4\pi/5</math> is the only other solution between <math>0</math> and <math>2\pi\,</math>.
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<math>x=\frac{\pi }{5}</math>
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is a solution to the equation, and using the unit circle we can also draw the conclusion that
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<math>x=\pi -\frac{\pi }{5}=\frac{4\pi }{5}</math>
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is the only other solution between
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<math>0</math>
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and
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<math>\text{2}\pi </math>.
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[[Image:4_4_3_b.gif|center]]
[[Image:4_4_3_b.gif|center]]
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We obtain all solutions to the equation when we add integer multiples of
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We obtain all solutions to the equation when we add integer multiples of <math>2\pi\, </math>,
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<math>\text{2}\pi </math>,
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{{Displayed math||<math>x = \frac{\pi}{5} + 2n\pi\qquad\text{and}\qquad x = \frac{4\pi}{5} + 2n\pi\,,</math>}}
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<math>x=\frac{\pi }{5}+2n\pi </math>
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and
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<math>x=\frac{4\pi }{5}+2n\pi </math>
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where ''n'' is an arbitrary integer.
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where
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<math>n</math>
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is an arbitrary integer.
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Current revision

We see directly that \displaystyle x = \pi/5 is a solution to the equation, and using the unit circle we can also draw the conclusion that \displaystyle x = \pi - \pi/5 = 4\pi/5 is the only other solution between \displaystyle 0 and \displaystyle 2\pi\,.

We obtain all solutions to the equation when we add integer multiples of \displaystyle 2\pi\, ,

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

where n is an arbitrary integer.

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