Solution 4.4:5c

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For a fixed value of
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For a fixed value of ''u'', an equality of the form
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<math>u</math>, an equality of the form
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{{Displayed math||<math>\cos u=\cos v</math>}}
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<math>\cos u=\cos v</math>
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is satisfied by two angles ''v'' in the unit circle,
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is satisfied by two angles
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<math>v</math>
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in the unit circle:
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<math>v=u</math>
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and
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<math>v=-u</math>
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{{Displayed math||<math>v=u\qquad\text{and}\qquad v=-u\,\textrm{.}</math>}}
[[Image:4_4_5_c.gif|center]]
[[Image:4_4_5_c.gif|center]]
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This means that all angles
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This means that all angles ''v'' which satisfy the equality are
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<math>v</math>
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which satisfy the equality are
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{{Displayed math||<math>v=u+2n\pi\qquad\text{and}\qquad v=-u+2n\pi\,,</math>}}
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<math>v=u+2n\pi </math>
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where ''n'' is an arbitrary integer.
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and
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<math>v=-u+2n\pi </math>
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where
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<math>n\text{ }</math>
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is an arbitrary integer.
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Therefore, the equation
Therefore, the equation
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{{Displayed math||<math>\cos 5x=\cos (x+\pi/5)</math>}}
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<math>\cos 5x=\cos \left( x+{\pi }/{5}\; \right)</math>
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has the solutions
has the solutions
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{{Displayed math||<math>\left\{\begin{align} 5x&=x+\frac{\pi}{5}+2n\pi\quad\text{or}\\[5pt] 5x &= -x-\frac{\pi}{5}+2n\pi\,\textrm{.}\end{align}\right.</math>}}
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<math>5x=x+\frac{\pi }{5}+2n\pi </math>
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or
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<math>5x=-x-\frac{\pi }{5}+2n\pi </math>
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If we collect
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If we collect ''x'' onto one side, we end up with
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<math>x\text{ }</math>
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onto one side, we end up with
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{{Displayed math||<math>\left\{\begin{align}
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x &= \frac{\pi}{20} + \frac{n\pi}{2}\,,\\[5pt]
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x &= -\frac{\pi }{30}+\frac{n\pi}{3}\,,
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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where ''n'' is an arbitrary integer.
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x=\frac{\pi }{20}+\frac{1}{2}n\pi \\
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x=-\frac{\pi }{30}+\frac{1}{3}n\pi \\
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\end{array} \right.</math>
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(
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<math>n\text{ }</math>
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an arbitrary integer).
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Current revision

For a fixed value of u, an equality of the form

\displaystyle \cos u=\cos v

is satisfied by two angles v in the unit circle,

\displaystyle v=u\qquad\text{and}\qquad v=-u\,\textrm{.}

This means that all angles v which satisfy the equality are

\displaystyle v=u+2n\pi\qquad\text{and}\qquad v=-u+2n\pi\,,

where n is an arbitrary integer.

Therefore, the equation

\displaystyle \cos 5x=\cos (x+\pi/5)

has the solutions

\displaystyle \left\{\begin{align} 5x&=x+\frac{\pi}{5}+2n\pi\quad\text{or}\\[5pt] 5x &= -x-\frac{\pi}{5}+2n\pi\,\textrm{.}\end{align}\right.

If we collect x onto one side, we end up with

\displaystyle \left\{\begin{align}

x &= \frac{\pi}{20} + \frac{n\pi}{2}\,,\\[5pt] x &= -\frac{\pi }{30}+\frac{n\pi}{3}\,, \end{align}\right.

where n is an arbitrary integer.

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