Solution 4.4:2f

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (12:44, 13 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
Using the unit circle shows that the equation
+
Using the unit circle shows that the equation <math>\cos 3x = -1/\!\sqrt{2}</math>
-
<math>\text{cos 3}x=-\frac{1}{\sqrt{2}}</math>
+
has two solutions for <math>0\le 3x\le 2\pi\,</math>,
-
has two solutions for
+
-
<math>0\le \text{3}x\le \text{2}\pi </math>,
+
-
 
+
{{Displayed math||<math>3x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\qquad\text{and}\qquad 3x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\,\textrm{.}</math>}}
-
<math>3x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}</math>
+
-
and
+
-
<math>3x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}</math>
+
[[Image:4_4_2_f.gif|center]]
[[Image:4_4_2_f.gif|center]]
-
We obtain the other solutions by adding multiples of
+
We obtain the other solutions by adding multiples of <math>2\pi</math>,
-
<math>2\pi </math>,
+
-
 
+
-
 
+
-
<math>3x=\frac{3\pi }{4}+2n\pi </math>
+
-
and
+
-
<math>3x=\frac{5\pi }{4}+2n\pi </math>
+
 +
{{Displayed math||<math>3x = \frac{3\pi}{4} + 2n\pi\qquad\text{and}\qquad 3x = \frac{5\pi}{4} + 2n\pi\,,</math>}}
i.e.
i.e.
 +
{{Displayed math||<math>x = \frac{\pi}{4} + \frac{2}{3}n\pi\qquad\text{and}\qquad x = \frac{5\pi}{12} + \frac{2}{3}n\pi\,,</math>}}
-
<math>x=\frac{\pi }{4}+\frac{2}{3}n\pi </math>
+
where ''n'' is an arbitrary integer.
-
and
+
-
<math>x=\frac{5\pi }{12}+\frac{2}{3}n\pi </math>
+
-
 
+
-
 
+
-
where
+
-
<math>n</math>
+
-
is an arbitrary integer.
+

Current revision

Using the unit circle shows that the equation \displaystyle \cos 3x = -1/\!\sqrt{2} has two solutions for \displaystyle 0\le 3x\le 2\pi\,,

\displaystyle 3x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\qquad\text{and}\qquad 3x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\,\textrm{.}

We obtain the other solutions by adding multiples of \displaystyle 2\pi,

\displaystyle 3x = \frac{3\pi}{4} + 2n\pi\qquad\text{and}\qquad 3x = \frac{5\pi}{4} + 2n\pi\,,

i.e.

\displaystyle x = \frac{\pi}{4} + \frac{2}{3}n\pi\qquad\text{and}\qquad x = \frac{5\pi}{12} + \frac{2}{3}n\pi\,,

where n is an arbitrary integer.

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.4:2f"