Solution 4.4:2f

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Using the unit circle shows that the equation
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<center> [[Image:4_4_2f.gif]] </center>
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<math>\text{cos 3}x=-\frac{1}{\sqrt{2}}</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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<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>
[[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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<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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i.e.
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<math>x=\frac{\pi }{4}+\frac{2}{3}n\pi </math>
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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.

Revision as of 08:57, 1 October 2008

Using the unit circle shows that the equation \displaystyle \text{cos 3}x=-\frac{1}{\sqrt{2}} has two solutions for \displaystyle 0\le \text{3}x\le \text{2}\pi ,


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

We obtain the other solutions by adding multiples of \displaystyle 2\pi ,


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


i.e.


\displaystyle x=\frac{\pi }{4}+\frac{2}{3}n\pi and \displaystyle x=\frac{5\pi }{12}+\frac{2}{3}n\pi


where \displaystyle n is an arbitrary integer.

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