Solution 4.4:5a

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If we consider for a moment the equality
If we consider for a moment the equality
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{{Displayed math||<math>\sin u = \sin v</math>|(*)}}
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<math>\sin u=\sin v\quad \quad \quad (*)</math>
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where ''u'' has a fixed value, there are usually two angles ''v'' in the unit circle which ensure that the equality holds,
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{{Displayed math||<math>v=u\qquad\text{and}\qquad v=\pi-u\,\textrm{.}</math>}}
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[[Image:4_4_5_a.gif||center]]
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where
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(The only exception is when <math>u = \pi/2</math> or <math>u=3\pi/2</math>, in which case <math>u</math> and <math>\pi-u</math> correspond to the same direction and there is only one angle ''v'' which satisfies the equality.)
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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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We obtain all the angles ''v'' which satisfy (*) by adding multiples of <math>2\pi</math>,
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<math>v=u</math>
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{{Displayed math||<math>v = u+2n\pi\qquad\text{and}\qquad v = \pi-u+2n\pi\,,</math>}}
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and
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<math>v=\pi -u</math>
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where ''n'' is an arbitrary integer.
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[[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
If we now go back to our equation
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{{Displayed math||<math>\sin 3x = \sin x</math>}}
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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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the reasoning above shows that the equation is only satisfied when
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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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If we make
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{{Displayed math||<math>3x = x+2n\pi\qquad\text{or}\qquad 3x = \pi-x+2n\pi\,\textrm{.}</math>}}
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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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If we make ''x'' the subject of each equation, we obtain the full solution to the equation,
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<math>\left\{ \begin{array}{*{35}l}
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{{Displayed math||<math>\left\{\begin{align}
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x=0+n\pi \\
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x &= 0+n\pi\,,\\[5pt]
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x=\frac{\pi }{4}+\frac{1}{2}n\pi \\
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x &= \frac{\pi}{4}+\frac{n\pi}{2}\,\textrm{.}
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\end{array} \right.</math>
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\end{align}\right.</math>}}

Current revision

If we consider for a moment the equality

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

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

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

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

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

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

where n 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\qquad\text{or}\qquad 3x = \pi-x+2n\pi\,\textrm{.}

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

\displaystyle \left\{\begin{align}

x &= 0+n\pi\,,\\[5pt] x &= \frac{\pi}{4}+\frac{n\pi}{2}\,\textrm{.} \end{align}\right.

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