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 | + | 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 “1” in the numerator of the left-hand side with <math>\sin^2\!x + \cos^2\!x</math> |
- | + | ||
- | + | ||
- | <math>\ | + | |
using the Pythagorean identity. This means that the equation's left-hand side can be written as | using the Pythagorean identity. This means that the equation's left-hand side can be written as | ||
- | + | {{Displayed math||<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>}} | |
- | <math>\frac{1}{\cos ^{2}x}=\frac{\cos ^ | + | |
- | + | ||
and the expression is then completely expressed in terms of tan x, | and the expression is then completely expressed in terms of tan x, | ||
+ | {{Displayed math||<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}} | ||
- | + | If we substitute <math>t=\tan x</math>, we see that we have a quadratic equation in ''t'', which, after simplifying, becomes <math>t^2+t=0</math> and has roots <math>t=0</math> and <math>t=-1</math>. There are therefore two possible values for | |
- | + | <math>\tan x</math>, <math>\tan x=0</math> or <math>\tan x=-1\,</math>. The first equality is satisfied when <math>x=n\pi</math> for all integers ''n'', and the second when <math>x=3\pi/4+n\pi\,</math>. | |
- | + | ||
- | If we substitute | + | |
- | <math>t=\tan x</math> | + | |
- | , we see that we have a | + | |
- | + | ||
- | , which, after simplifying, becomes | + | |
- | <math>t^ | + | |
- | and has roots | + | |
- | <math>t=0</math> | + | |
- | and | + | |
- | <math>t=- | + | |
- | <math>\tan x</math>, | + | |
- | <math>\tan x=0</math> | + | |
- | + | ||
- | <math>\tan x=-1</math> | + | |
- | The first equality is satisfied when | + | |
- | <math>x=n\pi </math> | + | |
- | for all integers | + | |
- | + | ||
- | <math>x= | + | |
The complete solution of the equation is | The complete solution of the equation is | ||
- | + | {{Displayed math||<math>\left\{\begin{align} | |
- | <math>\left\{ \begin{ | + | x &= n\pi\,,\\[5pt] |
- | x=n\pi | + | x &= \frac{3\pi}{4}+n\pi\,, |
- | x=\frac{3\pi }{4}+n\pi | + | \end{align}\right.</math>}} |
- | \end{ | + | |
- | + | where ''n'' is an arbitrary integer. | |
- | + | ||
- | an arbitrary integer | + |
Version vom 08:27, 14. 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 “1” in the numerator of the left-hand side with \displaystyle \sin^2\!x + \cos^2\!x using the Pythagorean identity. This means that the equation's left-hand side can be written as
and the expression is then completely expressed in terms of tan x,
If we substitute \displaystyle t=\tan x, we see that we have a quadratic equation in t, which, after simplifying, becomes \displaystyle t^2+t=0 and has roots \displaystyle t=0 and \displaystyle t=-1. There are therefore two possible values for \displaystyle \tan x, \displaystyle \tan x=0 or \displaystyle \tan x=-1\,. The first equality is satisfied when \displaystyle x=n\pi for all integers n, and the second when \displaystyle x=3\pi/4+n\pi\,.
The complete solution of the equation is
where n is an arbitrary integer.