Solution 4.4:6b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.4:6b moved to Solution 4.4:6b: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
After moving the terms over to the left-hand side, so that
-
<center> [[Image:4_4_6b-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\sqrt{2}\sin x\cos x-\cos x=0</math>
-
<center> [[Image:4_4_6b-2(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
we see that we can take out a common factor
 +
<math>\text{cos }x</math>,
 +
 
 +
 
 +
<math>\cos x\left( \sqrt{2}\sin x-1 \right)=0</math>
 +
 
 +
 
 +
and that the equation is only satisfied if at least one of the factors,
 +
<math>\text{cos }x</math>
 +
or
 +
<math>\sqrt{2}\text{sin }x-\text{1}</math>
 +
is zero. Thus, there are two cases:
 +
 
 +
 
 +
<math>\text{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
 +
 
 +
 
 +
<math>x=\frac{\pi }{2}+2n\pi </math>
 +
and
 +
<math>x=\frac{3\pi }{2}+2n\pi </math>
 +
 
 +
 
 +
where
 +
<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
 +
 
 +
 +
<math>x=\frac{\pi }{2}+n\pi </math>
 +
(
 +
<math>n</math>
 +
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
 +
 
 +
 
 +
<math>\left\{ \begin{array}{*{35}l}
 +
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).

Revision as of 11:41, 1 October 2008

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


\displaystyle \sqrt{2}\sin x\cos x-\cos x=0


we see that we can take out a common factor \displaystyle \text{cos }x,


\displaystyle \cos x\left( \sqrt{2}\sin x-1 \right)=0


and that the equation is only satisfied if at least one of the factors, \displaystyle \text{cos }x or \displaystyle \sqrt{2}\text{sin }x-\text{1} is zero. Thus, there are two cases:


\displaystyle \text{cos }x=0: This basic equation has solutions \displaystyle x={\pi }/{2}\; and \displaystyle x=3{\pi }/{2}\; in the unit circle, and from this we see that the general solution is


\displaystyle x=\frac{\pi }{2}+2n\pi and \displaystyle x=\frac{3\pi }{2}+2n\pi


where \displaystyle n\text{ } is an arbitrary integer. Because the angles \displaystyle {\pi }/{2}\; and \displaystyle 3{\pi }/{2}\; differ by \displaystyle \pi , the solutions can be summarized as


\displaystyle x=\frac{\pi }{2}+n\pi ( \displaystyle n an arbitrary integer).

√ \displaystyle \sqrt{2}\text{sin }x-\text{1}=0

if we rearrange the equation, we obtain the basic equation as

\displaystyle \text{sin }x\text{ }={1}/{\sqrt{2}}\;, which has the solutions \displaystyle x={\pi }/{4}\; and \displaystyle x=3{\pi }/{4}\; in the unit circle and hence the general solution


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


where \displaystyle n\text{ } can arbitrary integer.

All in all, the original equation has the solutions


\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{4}+2n\pi \\ x=\frac{\pi }{2}+n\pi \\ x=\frac{3\pi }{4}+2n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer).

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