Lösung 4.4:8c
Aus Online Mathematik Brückenkurs 1
When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “ \displaystyle \text{1} ” in the numerator of the left-hand side with \displaystyle \text{sin}^{\text{2}}x+\text{cos}^{\text{2}}x\text{ } using the Pythagorean identity. This means that the equation's left-hand side can be written as
\displaystyle \frac{1}{\cos ^{2}x}=\frac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=1+\frac{\sin ^{2}x}{\cos ^{2}x}=1+\tan ^{2}x
and the expression is then completely expressed in terms of tan x,
\displaystyle 1+\tan ^{2}x=1-\tan x
If we substitute
\displaystyle t=\tan x
, we see that we have a second-degree equation in
\displaystyle t
, which, after simplifying, becomes
\displaystyle t^{\text{2}}\text{ }+t=0
and has roots
\displaystyle t=0
and
\displaystyle t=-\text{1}. There are therefore two possible values for
\displaystyle \tan x,
\displaystyle \tan x=0
tan x =0 or
\displaystyle \tan x=-1
The first equality is satisfied when
\displaystyle x=n\pi
for all integers
\displaystyle n, and the second when
\displaystyle x=\frac{3\pi }{4}+n\pi .
The complete solution of the equation is
\displaystyle \left\{ \begin{array}{*{35}l}
x=n\pi \\
x=\frac{3\pi }{4}+n\pi \\
\end{array} \right.
(
\displaystyle n
an arbitrary integer).