Processing Math: 84%
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Solution 4.4:6b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:38, 13 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
After moving the terms over to the left-hand side, so that
After moving the terms over to the left-hand side, so that
 +
{{Displayed math||<math>\sqrt{2}\sin x\cos x-\cos x=0</math>}}
-
<math>\sqrt{2}\sin x\cos x-\cos x=0</math>
+
we see that we can take out a common factor <math>\cos x</math>,
 +
{{Displayed math||<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}
-
we see that we can take out a common factor
+
and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
-
<math>\text{cos }x</math>,
+
-
<math>\cos x\left( \sqrt{2}\sin x-1 \right)=0</math>
+
<math>\cos x=0:</math>
 +
This basic equation has solutions <math>x=\pi/2</math> and <math>x=3\pi/2</math> in the unit circle, and from this we see that the general solution is
-
and that the equation is only satisfied if at least one of the factors,
+
{{Displayed math||<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
-
<math>\text{cos }x</math>
+
-
or
+
-
<math>\sqrt{2}\text{sin }x-\text{1}</math>
+
-
is zero. Thus, there are two cases:
+
 +
where ''n'' is an arbitrary integer. Because the angles <math>\pi/2</math> and
 +
<math>3\pi/2</math> differ by <math>\pi</math>, the solutions can be summarized as
-
<math>\text{cos }x=0</math>: This basic equation has solutions
+
{{Displayed math||<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
-
<math>x={\pi }/{2}\;</math>
+
-
and
+
-
<math>x=3{\pi }/{2}\;</math>
+
-
in the unit circle, and from this we see that the general solution is
+
 +
where ''n'' is an arbitrary integer.
-
<math>x=\frac{\pi }{2}+2n\pi </math>
 
-
and
 
-
<math>x=\frac{3\pi }{2}+2n\pi </math>
 
 +
<math>\sqrt{2}\sin x - 1 = 0:</math>
-
where
+
If we rearrange the equation, we obtain the basic equation as <math>\sin x = 1/\!\sqrt{2}</math>, which has the solutions <math>x=\pi/4</math> and <math>x=3\pi /4</math> in the unit circle and hence the general solution
-
<math>n\text{ }</math>
+
-
is an arbitrary integer. Because the angles
+
-
<math>{\pi }/{2}\;</math>
+
-
and
+
-
<math>3{\pi }/{2}\;</math>
+
-
differ by
+
-
<math>\pi </math>, the solutions can be summarized as
+
-
+
{{Displayed math||<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
-
<math>x=\frac{\pi }{2}+n\pi </math>
+
-
(
+
-
<math>n</math>
+
-
an arbitrary integer).
+
-
√
+
where ''n'' is an arbitrary integer.
-
<math>\sqrt{2}\text{sin }x-\text{1}=0</math>
+
-
: if we rearrange the equation, we obtain the basic equation as
+
-
<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the solutions
+
-
<math>x={\pi }/{4}\;</math>
+
-
and
+
-
<math>x=3{\pi }/{4}\;</math>
+
-
in the unit circle and hence the general solution
+
- 
-
<math>x=\frac{\pi }{4}+2n\pi </math>
 
-
and
 
-
<math>x=\frac{3\pi }{4}+2n\pi </math>
 
- 
- 
-
where
 
-
<math>n\text{ }</math>
 
-
can arbitrary integer.
 
All in all, the original equation has the solutions
All in all, the original equation has the solutions
 +
{{Displayed math||<math>\left\{\begin{align}
 +
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
 +
x &= \frac{\pi}{2}+n\pi\,,\\[5pt]
 +
x &= \frac{3\pi}{4}+2n\pi\,,
 +
\end{align}\right.</math>}}
-
<math>\left\{ \begin{array}{*{35}l}
+
where ''n'' is an arbitrary integer.
-
x=\frac{\pi }{4}+2n\pi \\
+
-
x=\frac{\pi }{2}+n\pi \\
+
-
x=\frac{3\pi }{4}+2n\pi \\
+
-
\end{array} \right.</math>
+
-
(
+
-
<math>n\text{ }</math>
+
-
an arbitrary integer).
+

Current revision

After moving the terms over to the left-hand side, so that

2sinxcosx−cosx=0 

we see that we can take out a common factor cosx,

cosx(2sinx−1)=0 

and that the equation is only satisfied if at least one of the factors cosx or 2sinx−1  is zero. Thus, there are two cases:


cosx=0:

This basic equation has solutions x=2 and x=32 in the unit circle, and from this we see that the general solution is

x=2+2nandx=23+2n

where n is an arbitrary integer. Because the angles 2 and 32 differ by , the solutions can be summarized as

x=2+n

where n is an arbitrary integer.


2sinx−1=0: 

If we rearrange the equation, we obtain the basic equation as sinx=12 , which has the solutions x=4 and \displaystyle x=3\pi /4 in the unit circle and hence the general solution

\displaystyle x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,

where n is an arbitrary integer.


All in all, the original equation has the solutions

\displaystyle \left\{\begin{align}

x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{\pi}{2}+n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.4:6b"