Lösung 4.4:8c
Aus Online Mathematik Brückenkurs 1
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- | {{ | + | 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 “ |
- | < | + | <math>\text{1}</math> |
- | { | + | ” in the numerator of the left-hand side with |
- | {{ | + | <math>\text{sin}^{\text{2}}x+\text{cos}^{\text{2}}x\text{ }</math> |
- | < | + | using the Pythagorean identity. This means that the equation's left-hand side can be written as |
- | {{ | + | |
+ | |||
+ | <math>\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</math> | ||
+ | |||
+ | |||
+ | and the expression is then completely expressed in terms of tan x, | ||
+ | |||
+ | |||
+ | <math>1+\tan ^{2}x=1-\tan x</math> | ||
+ | |||
+ | |||
+ | If we substitute | ||
+ | <math>t=\tan x</math> | ||
+ | , we see that we have a second-degree equation in | ||
+ | <math>t</math> | ||
+ | , which, after simplifying, becomes | ||
+ | <math>t^{\text{2}}\text{ }+t=0</math> | ||
+ | and has roots | ||
+ | <math>t=0</math> | ||
+ | and | ||
+ | <math>t=-\text{1}</math>. There are therefore two possible values for | ||
+ | <math>\tan x</math>, | ||
+ | <math>\tan x=0</math> | ||
+ | tan x =0 or | ||
+ | <math>\tan x=-1</math> | ||
+ | The first equality is satisfied when | ||
+ | <math>x=n\pi </math> | ||
+ | for all integers | ||
+ | <math>n</math>, and the second when | ||
+ | <math>x=\frac{3\pi }{4}+n\pi </math>. | ||
+ | |||
+ | The complete solution of the equation is | ||
+ | |||
+ | |||
+ | <math>\left\{ \begin{array}{*{35}l} | ||
+ | x=n\pi \\ | ||
+ | x=\frac{3\pi }{4}+n\pi \\ | ||
+ | \end{array} \right.</math> | ||
+ | ( | ||
+ | <math>n</math> | ||
+ | an arbitrary integer). |
Version vom 08:14, 2. Okt. 2008
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).