Lösung 4.4:7a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
-
{{Displayed math||<math>2t^2 + t = 1</math>}}
+
{{Abgesetzte Formel||<math>2t^2 + t = 1</math>}}
and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,
and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
t^2 + \frac{1}{2}t - \frac{1}{2}
t^2 + \frac{1}{2}t - \frac{1}{2}
&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
Zeile 15: Zeile 15:
and then obtain the equation
and then obtain the equation
-
{{Displayed math||<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}
+
{{Abgesetzte Formel||<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}
which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.
which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.
-
{{Displayed math||<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}
Zeile 29: Zeile 29:
This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is
This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is
-
{{Displayed math||<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.
Zeile 38: Zeile 38:
The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore
The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore
-
{{Displayed math||<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}
+
{{Abgesetzte Formel||<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}
where ''n'' is an arbitrary integer.
where ''n'' is an arbitrary integer.
Zeile 45: Zeile 45:
All of the solution to the equation are given by
All of the solution to the equation are given by
-
{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
x &= \pi/6+2n\pi\,,\\[5pt]
x &= \pi/6+2n\pi\,,\\[5pt]
x &= 5\pi/6+2n\pi\,,\\[5pt]
x &= 5\pi/6+2n\pi\,,\\[5pt]

Version vom 09:00, 22. Okt. 2008

If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \sin x and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \sin x, instead of trying to solve for \displaystyle x directly.

If we write \displaystyle t = \sin x and treat \displaystyle t as a new unknown variable, the equation becomes

\displaystyle 2t^2 + t = 1

and it is expressed completely in terms of \displaystyle t. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,

\displaystyle \begin{align}

t^2 + \frac{1}{2}t - \frac{1}{2} &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt] &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16} \end{align}

and then obtain the equation

\displaystyle \Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}

which has the solutions \displaystyle t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}, i.e.

\displaystyle t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}


Because \displaystyle t=\sin x, this means that the values of x that satisfy the equation in the exercise will necessarily satisfy one of the basic equations \displaystyle \sin x = \tfrac{1}{2} or \displaystyle \sin x = -1\,.


\displaystyle \sin x = \frac{1}{2}:

This equation has the solutions \displaystyle x = \pi/6 and \displaystyle x = \pi - \pi/6 = 5\pi/6 in the unit circle and the general solution is

\displaystyle x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,

where n is an arbitrary integer.


\displaystyle \sin x = -1:

The equation has only one solution \displaystyle x = 3\pi/2 in the unit circle, and the general solution is therefore

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

where n is an arbitrary integer.


All of the solution to the equation are given by

\displaystyle \left\{\begin{align}

x &= \pi/6+2n\pi\,,\\[5pt] x &= 5\pi/6+2n\pi\,,\\[5pt] x &= 3\pi/2+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:7a“