Solution 4.4:5a

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If we consider for a moment the equality
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<center> [[Image:4_4_5a-1(2).gif]] </center>
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<math>\sin u=\sin v\quad \quad \quad (*)</math>
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<center> [[Image:4_4_5a-2(2).gif]] </center>
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where
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<math>u</math>
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has a fixed value, there are usually two angles
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<math>v</math>
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in the unit circle which ensure that the equality holds:
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<math>v=u</math>
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and
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<math>v=\pi -u</math>
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[[Image:4_4_5_a.gif]]
[[Image:4_4_5_a.gif]]
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(The only exception is when
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<math>u={\pi }/{2}\;\text{ }</math>
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or
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<math>u=3{\pi }/{2}\;\text{ }</math>, in which case
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<math>u</math>
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and
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<math>\pi -u\text{ }</math>
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correspond to the same direction and there is only one angle
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<math>v</math>
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which satisfies the equality.)
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We obtain all the angles
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<math>v</math>
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which satisfy (*) by adding multiples of
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<math>\text{2}\pi </math>,
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<math>v=u+2n\pi </math>
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and
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<math>v=\pi -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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If we now go back to our equation
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<math>\sin 3x=\sin x</math>
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the reasoning above shows that the equation is only satisfied when
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 +
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<math>3x=x+2n\pi </math>
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or
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<math>3x=\pi -x+2n\pi </math>
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 +
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If we make
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<math>x</math>
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the subject of each equation, we obtain the full solution to the equation:
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 +
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<math>\left\{ \begin{array}{*{35}l}
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x=0+n\pi \\
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x=\frac{\pi }{4}+\frac{1}{2}n\pi \\
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\end{array} \right.</math>

Revision as of 10:55, 1 October 2008

If we consider for a moment the equality


\displaystyle \sin u=\sin v\quad \quad \quad (*)


where \displaystyle u has a fixed value, there are usually two angles \displaystyle v in the unit circle which ensure that the equality holds:


\displaystyle v=u and \displaystyle v=\pi -u


Image:4_4_5_a.gif


(The only exception is when \displaystyle u={\pi }/{2}\;\text{ } or \displaystyle u=3{\pi }/{2}\;\text{ }, in which case \displaystyle u and \displaystyle \pi -u\text{ } correspond to the same direction and there is only one angle \displaystyle v which satisfies the equality.)

We obtain all the angles \displaystyle v which satisfy (*) by adding multiples of \displaystyle \text{2}\pi ,


\displaystyle v=u+2n\pi and \displaystyle v=\pi -u+2n\pi


where \displaystyle n\text{ } is an arbitrary integer.

If we now go back to our equation


\displaystyle \sin 3x=\sin x


the reasoning above shows that the equation is only satisfied when


\displaystyle 3x=x+2n\pi or \displaystyle 3x=\pi -x+2n\pi


If we make \displaystyle x the subject of each equation, we obtain the full solution to the equation:


\displaystyle \left\{ \begin{array}{*{35}l} x=0+n\pi \\ x=\frac{\pi }{4}+\frac{1}{2}n\pi \\ \end{array} \right.

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