Lösung 4.4:6b
Aus Online Mathematik Brückenkurs 1
After moving the terms over to the left-hand side, so that
we see that we can take out a common factor \displaystyle \cos x,
and that the equation is only satisfied if at least one of the factors \displaystyle \cos x or \displaystyle \sqrt{2}\sin x - 1 is zero. Thus, there are two cases:
\displaystyle \cos x=0:
This basic equation has solutions \displaystyle x=\pi/2 and \displaystyle x=3\pi/2 in the unit circle, and from this we see that the general solution is
where n is an arbitrary integer. Because the angles \displaystyle \pi/2 and \displaystyle 3\pi/2 differ by \displaystyle \pi, the solutions can be summarized as
where n is an arbitrary integer.
\displaystyle \sqrt{2}\sin x - 1 = 0:
If we rearrange the equation, we obtain the basic equation as \displaystyle \sin x = 1/\!\sqrt{2}, which has the solutions \displaystyle x=\pi/4 and \displaystyle x=3\pi /4 in the unit circle and hence the general solution
where n is an arbitrary integer.
All in all, the original equation has the solutions
where n is an arbitrary integer.
