Solution 4.4:2d

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Current revision (14:32, 10 October 2008) (edit) (undo)
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Apart from the fact that there is a
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Apart from the fact that there is a <math>5x</math>, this is a normal trigonometric equation of the type <math>\sin y = a\,</math>. If we are only interested in solutions which satisfy <math>0\le 5x\le 2\pi</math>, then a sketch of the unit circle shows that there are two such solutions, <math>5x = \pi/4</math> and the reflectionally symmetric solution <math>5x = \pi - \pi/4 = 3\pi/4\,</math>.
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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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<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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All of the equation's solutions are obtained from all values of <math>5x</math> which differ by a multiple of <math>2\pi</math> 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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{{Displayed math||<math>5x = \frac{\pi}{4} + 2n\pi\qquad\text{and}\qquad 5x = \frac{3\pi}{4} + 2n\pi\,,</math>}}
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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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where ''n'' is an arbitrary integer.
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<math>\text{5}</math>, we obtain the solutions expressed in terms of x alone:
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If we divide both of these by 5, 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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{{Displayed math||<math>x = \frac{\pi}{20} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{3\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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where ''n'' is an arbitrary integer.
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<math>n</math>
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is an arbitrary integer.
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Current revision

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

All of the equation's solutions are obtained from all values of \displaystyle 5x which differ by a multiple of \displaystyle 2\pi from either of these two solutions,

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

where n is an arbitrary integer.

If we divide both of these by 5, we obtain the solutions expressed in terms of x alone,

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

where n is an arbitrary integer.

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