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 can be written in terms of \displaystyle \cos x,
or, in rearranged form,
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
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
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
where n is an arbitrary integer.
