Lösung 4.4:7b

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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>,
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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{{Displayed math||<math>2(1-\cos^2\!x) - 3\cos x = 0\,,</math>}}
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{{Abgesetzte Formel||<math>2(1-\cos^2\!x) - 3\cos x = 0\,,</math>}}
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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{{Abgesetzte Formel||<math>2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}</math>}}
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
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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{{Displayed math||<math>2t^2+3t-2 = 0</math>}}
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{{Abgesetzte Formel||<math>2t^2+3t-2 = 0</math>}}
and this quadratic equation has the solutions <math>t=\tfrac{1}{2}</math> and
and this quadratic equation has the solutions <math>t=\tfrac{1}{2}</math> and
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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
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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{{Displayed math||<math>x=\pm \frac{\pi}{3}+2n\pi\qquad</math>(''n'' is an arbitrary integer),}}
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{{Abgesetzte Formel||<math>x=\pm \frac{\pi}{3}+2n\pi\qquad</math>(''n'' is an arbitrary integer),}}
whilst the equation <math>\cos x = -2</math> has no solutions at all (the values of cosine lie between -1 and 1).
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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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>}}
+
{{Abgesetzte Formel||<math>x = \pm\frac{\pi}{3} + 2n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.

Version vom 09:00, 22. Okt. 2008

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.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:7b“