Solution 4.4:6c

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If we use the trigonometric relation
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If we use the trigonometric relation <math>\sin (-x) = -\sin x</math>, the equation can be rewritten as
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<math>\text{sin }\left( -x \right)=-\text{sin }x</math>, the equation can be rewritten as
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<math>\sin 2x=\sin \left( -x \right)</math>
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{{Displayed math||<math>\sin 2x = \sin (-x)\,\textrm{.}</math>}}
In exercise 4.4:5a, we saw that an equality of the type
In exercise 4.4:5a, we saw that an equality of the type
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{{Displayed math||<math>\sin u = \sin v</math>}}
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<math>\sin u=\sin v\quad </math>
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is satisfied if
is satisfied if
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{{Displayed math||<math>u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,</math>}}
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<math>u=v+2n\pi </math>
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where ''n'' is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
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or
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<math>u=\pi -v+2n\pi </math>
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where
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<math>n\text{ }</math>
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is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
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<math>2x=-x+2n\pi </math>
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or
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<math>2x=\pi -\left( -x \right)+2n\pi </math>
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{{Displayed math||<math>2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,</math>}}
i.e.
i.e.
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{{Displayed math||<math>3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}</math>}}
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<math>3x=2n\pi </math>
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The solutions to the equation are thus
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or
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<math>x=\pi +2n\pi </math>
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{{Displayed math||<math>\left\{\begin{align}
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x &= \frac{2n\pi}{3}\,,\\[5pt]
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x &= \pi + 2n\pi\,,
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\end{align}\right.</math>}}
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The solutions to the equation are thus
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where ''n'' is an arbitrary integer.
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<math>\left\{ \begin{array}{*{35}l}
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x=\frac{2n\pi }{3} \\
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x=\pi +2n\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

If we use the trigonometric relation \displaystyle \sin (-x) = -\sin x, the equation can be rewritten as

\displaystyle \sin 2x = \sin (-x)\,\textrm{.}

In exercise 4.4:5a, we saw that an equality of the type

\displaystyle \sin u = \sin v

is satisfied if

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

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

\displaystyle 2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,

i.e.

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

The solutions to the equation are thus

\displaystyle \left\{\begin{align}

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

where n is an arbitrary integer.

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