Lösung 4.4:7b
Aus Online Mathematik Brückenkurs 1
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).
