Lösung 4.4:7b
Aus Online Mathematik Brückenkurs 1
K (Lösning 4.4:7b moved to Solution 4.4:7b: Robot: moved page) |
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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 written in terms of |
| - | < | + | <math>\cos x</math> |
| - | {{ | + | becomes |
| + | |||
| + | |||
| + | <math>2\left( 1-\cos ^{2}x \right)-3\cos x=0</math> | ||
| + | |||
| + | |||
| + | <math></math> | ||
| + | |||
| + | or, in rearranged form, | ||
| + | |||
| + | |||
| + | <math>2\cos ^{2}x+3\cos x-2=0</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 | ||
| + | <math>t</math>. Expressed in terms of | ||
| + | <math>t</math>, the equation is | ||
| + | |||
| + | |||
| + | <math>2t^{2}+3t-2=0</math> | ||
| + | |||
| + | |||
| + | and this second-degree equation has the solutions | ||
| + | <math>t=\frac{1}{2}</math> | ||
| + | and | ||
| + | <math>t=-2</math> | ||
| + | . | ||
| + | |||
| + | In terms of | ||
| + | <math>x</math>, this means that either | ||
| + | <math>\cos x=\frac{1}{2}</math> | ||
| + | or | ||
| + | <math>\text{cos }x=-\text{2}</math>. The first case occurs when | ||
| + | |||
| + | |||
| + | <math>x=\pm \frac{\pi }{3}+2n\pi </math> | ||
| + | ( | ||
| + | <math>n</math> | ||
| + | an arbitrary integer), | ||
| + | |||
| + | whilst the equation | ||
| + | <math>\text{cos }x=-\text{2 }</math> | ||
| + | has no solutions at all (the values of cosine lie between | ||
| + | <math>-\text{1 }</math> | ||
| + | and | ||
| + | <math>\text{1}</math> | ||
| + | ). | ||
| + | |||
| + | The answer is that the equation has the solutions | ||
| + | |||
| + | |||
| + | <math>x=\pm \frac{\pi }{3}+2n\pi </math> | ||
| + | ( | ||
| + | <math>n</math> | ||
| + | an arbitrary integer). | ||
Version vom 13:01, 1. Okt. 2008
If we use the Pythagorean identity and write \displaystyle \sin ^{2}x as \displaystyle 1-\cos ^{2}x, the whole equation written in terms of \displaystyle \cos x becomes
\displaystyle 2\left( 1-\cos ^{2}x \right)-3\cos x=0
\displaystyle
or, in rearranged form,
\displaystyle 2\cos ^{2}x+3\cos x-2=0
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
\displaystyle t. Expressed in terms of
\displaystyle t, the equation is
\displaystyle 2t^{2}+3t-2=0
and this second-degree equation has the solutions
\displaystyle t=\frac{1}{2}
and
\displaystyle t=-2
.
In terms of \displaystyle x, this means that either \displaystyle \cos x=\frac{1}{2} or \displaystyle \text{cos }x=-\text{2}. The first case occurs when
\displaystyle x=\pm \frac{\pi }{3}+2n\pi
(
\displaystyle n
an arbitrary integer),
whilst the equation \displaystyle \text{cos }x=-\text{2 } has no solutions at all (the values of cosine lie between \displaystyle -\text{1 } and \displaystyle \text{1} ).
The answer is that the equation has the solutions
\displaystyle x=\pm \frac{\pi }{3}+2n\pi
(
\displaystyle n
an arbitrary integer).
