Processing Math: Done
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 4.4:8a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-[[Bild: +[[Image:))
Current revision (08:12, 14 October 2008) (edit) (undo)
m
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
-
<center> [[Image:4_4_8a-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}</math>}}
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Image:4_4_8a-2(2).gif]] </center>
+
Then, we see that we can take a factor <math>\cos x</math> out of both terms,
-
{{NAVCONTENT_STOP}}
+
 
 +
{{Displayed math||<math>\cos x\,(2\sin x-\sqrt{2}) = 0</math>}}
 +
 
 +
and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.
 +
 
 +
 
 +
<math>\cos x = 0</math>:
 +
 
 +
This equation has the general solution
 +
 +
{{Displayed math||<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
 +
 
 +
 
 +
<math>2\sin x-\sqrt{2}=0</math>:
 +
 
 +
If we collect <math>\sin x</math> on the left-hand side, we obtain the equation
 +
<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
 +
 
 +
{{Displayed math||<math>\left\{\begin{align}
 +
x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
 +
x &= \frac{3\pi}{4}+2n\pi\,,
 +
\end{align}\right.</math>}}
 +
 
 +
where ''n'' is an arbitrary integer.
 +
 +
 
 +
The complete solution of the equation is
 +
 
 +
{{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>}}
 +
 
 +
where ''n'' is an arbitrary integer.

Current revision

If we use the formula for double angles, sin2x=2sinxcosx, and move all the terms over to the left-hand side, the equation becomes

2sinxcosx−2cosx=0. 

Then, we see that we can take a factor cosx out of both terms,

cosx(2sinx−2)=0 

and hence divide up the equation into two cases. The equation is satisfied either if cosx=0 or if 2sinx−2=0 .


cosx=0:

This equation has the general solution

x=2+n(n is an arbitrary integer).


2sinx−2=0 :

If we collect sinx on the left-hand side, we obtain the equation sinx=12 , which has the general solution

xx=4+2n=43+2n

where n is an arbitrary integer.


The complete solution of the equation is

xxx=4+2n=2+n=43+2n

where n is an arbitrary integer.

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.4:8a"