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Solution 4.4:2f

From Förberedande kurs i matematik 1

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Current revision (12:44, 13 October 2008) (edit) (undo)
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Using the unit circle shows that the equation
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Using the unit circle shows that the equation <math>\cos 3x = -1/\!\sqrt{2}</math>
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<math>\text{cos 3}x=-\frac{1}{\sqrt{2}}</math>
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has two solutions for <math>0\le 3x\le 2\pi\,</math>,
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has two solutions for
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<math>0\le \text{3}x\le \text{2}\pi </math>,
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{{Displayed math||<math>3x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\qquad\text{and}\qquad 3x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\,\textrm{.}</math>}}
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<math>3x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}</math>
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and
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<math>3x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}</math>
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[[Image:4_4_2_f.gif|center]]
[[Image:4_4_2_f.gif|center]]
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We obtain the other solutions by adding multiples of
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We obtain the other solutions by adding multiples of <math>2\pi</math>,
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<math>2\pi </math>,
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<math>3x=\frac{3\pi }{4}+2n\pi </math>
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and
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<math>3x=\frac{5\pi }{4}+2n\pi </math>
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{{Displayed math||<math>3x = \frac{3\pi}{4} + 2n\pi\qquad\text{and}\qquad 3x = \frac{5\pi}{4} + 2n\pi\,,</math>}}
i.e.
i.e.
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{{Displayed math||<math>x = \frac{\pi}{4} + \frac{2}{3}n\pi\qquad\text{and}\qquad x = \frac{5\pi}{12} + \frac{2}{3}n\pi\,,</math>}}
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<math>x=\frac{\pi }{4}+\frac{2}{3}n\pi </math>
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where ''n'' is an arbitrary integer.
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and
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<math>x=\frac{5\pi }{12}+\frac{2}{3}n\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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Current revision

Using the unit circle shows that the equation cos3x=−12  has two solutions for 03x2,

3x=2+4=43and3x=+4=45.

We obtain the other solutions by adding multiples of 2,

3x=43+2nand3x=45+2n

i.e.

x=4+32nandx=125+32n

where n is an arbitrary integer.

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