Lösung 4.4:7b
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Version vom 15:18, 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.
