Lösung 4.4:7a

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If we examine the equation, we see that
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If we examine the equation, we see that <math>x</math> only occurs as <math>\sin x</math> and it can therefore be appropriate to take an intermediary step and solve for <math>\sin x</math>, instead of trying to solve for <math>x</math> directly.
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<math>x</math>
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only occurs as
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<math>\text{sin }x\text{ }</math>
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and it can therefore be appropriate to take an intermediary step and solve for
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<math>\text{sin }x</math>, instead of trying to solve for
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<math>x</math>
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directly.
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If we write
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If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
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<math>t=\sin x</math>
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and treat
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<math>t</math>
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as a new unknown variable, the equation becomes
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{{Displayed math||<math>2t^2 + t = 1</math>}}
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<math>2t^{2}+t=1</math>
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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,
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when it is expressed completely in terms of
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<math>t</math>. This is a normal second-degree equation; after dividing by
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<math>\text{2}</math>, we complete the square on the left-hand side,
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<math>\begin{align}
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& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\
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& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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t^2 + \frac{1}{2}t - \frac{1}{2}
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&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
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&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16}
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\end{align}</math>}}
and then obtain the equation
and then obtain the equation
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{{Displayed math||<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}
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<math>\left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}</math>
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which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.
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which has the solutions
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<math>t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}</math>,
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i.e.
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<math>t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}</math>
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and
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<math>t=-\frac{1}{4}-\frac{3}{4}=-1</math>
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Because
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{{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>}}
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<math>t=\sin x</math>, this means that the values of
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<math>x</math>
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that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,
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<math>\text{sin }x=\frac{1}{2}</math>
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or
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<math>\text{sin }x=-\text{1}.</math>
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Because <math>t=\sin x</math>, this means that the values of ''x'' that satisfy the equation in the exercise will necessarily satisfy one of the basic equations <math>\sin x = \tfrac{1}{2}</math> or <math>\sin x = -1\,</math>.
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<math>\text{sin }x=\frac{1}{2}</math>: this equation has the solutions
 
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<math>x={\pi }/{6}\;</math>
 
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and
 
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<math>x=\pi -{\pi }/{6}\;=5{\pi }/{6}\;</math>
 
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in the unit circle and the general solution is
 
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<math>\sin x = \frac{1}{2}</math>:
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<math>x=\frac{\pi }{6}+2n\pi </math>
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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
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and
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<math>x=\frac{5\pi }{6}+2n\pi </math>
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{{Displayed math||<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}
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where
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where ''n'' is an arbitrary integer.
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<math>n\text{ }</math>
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is an arbitrary integer.
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<math>\text{sin }x=-\text{1}</math>: the equation has only one solution
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<math>\sin x = -1</math>:
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<math>x={3\pi }/{2}\;</math>
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in the unit circle, and the general solution is therefore
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The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore
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<math>x=\frac{3\pi }{2}+2n\pi </math>
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{{Displayed math||<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.
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where
 
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<math>n\text{ }</math>
 
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is an arbitrary integer.
 
All of the solution to the equation are given by
All of the solution to the equation are given by
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{{Displayed math||<math>\left\{\begin{align}
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x &= \pi/6+2n\pi\,,\\[5pt]
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x &= 5\pi/6+2n\pi\,,\\[5pt]
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x &= 3\pi/2+2n\pi\,,
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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where ''n'' is an arbitrary integer.
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x={\pi }/{6}\;+2n\pi \\
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x={5\pi }/{6}\;+2n\pi \\
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x={3\pi }/{2}\;+2n\pi \\
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\end{array} \right.</math>
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(
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<math>n\text{ }</math>
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an arbitrary integer)
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Version vom 07:32, 14. 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

Vorlage:Displayed math

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,

Vorlage:Displayed math

and then obtain the equation

Vorlage:Displayed math

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

Vorlage:Displayed math


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

Vorlage:Displayed math

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

Vorlage:Displayed math

where n is an arbitrary integer.


All of the solution to the equation are given by

Vorlage:Displayed math

where n is an arbitrary integer.

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