Solution 4.4:7b

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Current revision (07:44, 14 October 2008) (edit) (undo)
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If we use the Pythagorean identity and write
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If we use the Pythagorean identity and write <math>\sin^2\!x</math> as <math>1-\cos^2\!x</math>, the whole equation can be written in terms of <math>\cos x</math>,
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<math>\sin ^{2}x</math>
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as
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<math>1-\cos ^{2}x</math>, the whole equation written in terms of
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<math>\cos x</math>
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becomes
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{{Displayed math||<math>2(1-\cos^2\!x) - 3\cos x = 0\,,</math>}}
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<math>2\left( 1-\cos ^{2}x \right)-3\cos x=0</math>
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<math></math>
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or, in rearranged form,
or, in rearranged form,
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{{Displayed math||<math>2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}</math>}}
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<math>2\cos ^{2}x+3\cos x-2=0</math>
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With the equation expressed entirely in terms of <math>\cos x</math>, we can introduce a new unknown variable <math>t=\cos x</math> and solve the equation with respect to ''t''. Expressed in terms of ''t'', the equation is
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With the equation expressed entirely in terms of
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<math>\cos x</math>, we can introduce a new unknown variable
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<math>t=\cos x</math>
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and solve the equation with respect to
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<math>t</math>. Expressed in terms of
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<math>t</math>, the equation is
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<math>2t^{2}+3t-2=0</math>
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and this second-degree equation has the solutions
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{{Displayed math||<math>2t^2+3t-2 = 0</math>}}
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<math>t=\frac{1}{2}</math>
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and
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<math>t=-2</math>
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.
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In terms of
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and this quadratic equation has the solutions <math>t=\tfrac{1}{2}</math> and
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<math>x</math>, this means that either
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<math>t=-2\,</math>.
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<math>\cos x=\frac{1}{2}</math>
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or
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<math>\text{cos }x=-\text{2}</math>. The first case occurs when
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In terms of ''x'', this means that either <math>\cos x = \tfrac{1}{2}</math> or <math>\cos x = -2</math>. The first case occurs when
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<math>x=\pm \frac{\pi }{3}+2n\pi </math>
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{{Displayed math||<math>x=\pm \frac{\pi}{3}+2n\pi\qquad</math>(''n'' is an arbitrary integer),}}
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(
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<math>n</math>
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an arbitrary integer),
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whilst the equation
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whilst the equation <math>\cos x = -2</math> has no solutions at all (the values of cosine lie between -1 and 1).
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<math>\text{cos }x=-\text{2 }</math>
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has no solutions at all (the values of cosine lie between
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<math>-\text{1 }</math>
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and
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<math>\text{1}</math>
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).
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The answer is that the equation has the solutions
The answer is that the equation has the solutions
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{{Displayed math||<math>x = \pm\frac{\pi}{3} + 2n\pi\,,</math>}}
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<math>x=\pm \frac{\pi }{3}+2n\pi </math>
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where ''n'' is an arbitrary integer.
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(
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<math>n</math>
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an arbitrary integer).
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Current revision

If we use the Pythagorean identity and write \displaystyle \sin^2\!x as \displaystyle 1-\cos^2\!x, the whole equation can be written in terms of \displaystyle \cos x,

\displaystyle 2(1-\cos^2\!x) - 3\cos x = 0\,,

or, in rearranged form,

\displaystyle 2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}

With the equation expressed entirely in terms of \displaystyle \cos x, we can introduce a new unknown variable \displaystyle t=\cos x and solve the equation with respect to t. Expressed in terms of t, the equation is

\displaystyle 2t^2+3t-2 = 0

and this quadratic equation has the solutions \displaystyle t=\tfrac{1}{2} and \displaystyle t=-2\,.

In terms of x, this means that either \displaystyle \cos x = \tfrac{1}{2} or \displaystyle \cos x = -2. The first case occurs when

\displaystyle x=\pm \frac{\pi}{3}+2n\pi\qquad(n is an arbitrary integer),

whilst the equation \displaystyle \cos x = -2 has no solutions at all (the values of cosine lie between -1 and 1).

The answer is that the equation has the solutions

\displaystyle x = \pm\frac{\pi}{3} + 2n\pi\,,

where n is an arbitrary integer.

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