Aus Online Mathematik Brückenkurs 1

Version vom 10:07, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 4.4:7a moved to Solution 4.4:7a: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 12:49, 1. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	If we examine the equation, we see that
-	<center> [[Image:4_4_7a-1(2).gif]] </center>	+	<math>x</math>
-	{{NAVCONTENT_STOP}}	+	only occurs as
-	{{NAVCONTENT_START}}	+	<math>\text{sin }x\text{ }</math>
-	<center> [[Image:4_4_7a-2(2).gif]] </center>	+	and it can therefore be appropriate to take an intermediary step and solve for
-	{{NAVCONTENT_STOP}}	+	<math>\text{sin }x</math>, instead of trying to solve for
		+	<math>x</math>
		+	directly.
		+
		+	If we write
		+	<math>t=\sin x</math>
		+	and treat
		+	<math>t</math>
		+	as a new unknown variable, the equation becomes
		+
		+
		+	<math>2t^{2}+t=1</math>
		+
		+
		+	when it is expressed completely in terms of
		+	<math>t</math>. This is a normal second-degree equation; after dividing by
		+	<math>\text{2}</math>, we complete the square on the left-hand side,
		+
		+
		+	<math>\begin{align}
		+	& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\
		+	& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\
		+	\end{align}</math>
		+
		+
		+	and then obtain the equation
		+
		+
		+	<math>\left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}</math>
		+
		+
		+	which has the solutions
		+	<math>t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}</math>,
		+	i.e.
		+	<math>t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}</math>
		+	and
		+	<math>t=-\frac{1}{4}-\frac{3}{4}=-1</math>
		+
		+
		+	Because
		+	<math>t=\sin x</math>, this means that the values of
		+	<math>x</math>
		+	that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,
		+	<math>\text{sin }x=\frac{1}{2}</math>
		+	or
		+	<math>\text{sin }x=-\text{1}.</math>
		+
		+
		+
		+	<math>\text{sin }x=\frac{1}{2}</math>: 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
		+
		+
		+	<math>x=\frac{\pi }{6}+2n\pi </math>
		+	and
		+	<math>x=\frac{5\pi }{6}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer.
		+
		+
		+	<math>\text{sin }x=-\text{1}</math>: the equation has only one solution
		+	<math>x={3\pi }/{2}\;</math>
		+	in the unit circle, and the general solution is therefore
		+
		+
		+	<math>x=\frac{3\pi }{2}+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer.
		+
		+	All of the solution to the equation are given by
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x={\pi }/{6}\;+2n\pi \\
		+	x={5\pi }/{6}\;+2n\pi \\
		+	x={3\pi }/{2}\;+2n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n\text{ }</math>
		+	an arbitrary integer)

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Version vom 12:49, 1. Okt. 2008

If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \text{sin }x\text{ } and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \text{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

when it is expressed completely in terms of \displaystyle t. This is a normal second-degree equation; after dividing by \displaystyle \text{2}, we complete the square on the left-hand side,

\displaystyle \begin{align} & 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\ & =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\ \end{align}

and then obtain the equation

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

which has the solutions \displaystyle t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}, i.e. \displaystyle t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2} and \displaystyle t=-\frac{1}{4}-\frac{3}{4}=-1

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

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

where \displaystyle n\text{ } is an arbitrary integer.

\displaystyle \text{sin }x=-\text{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 \displaystyle n\text{ } is an arbitrary integer.

All of the solution to the equation are given by

\displaystyle \left\{ \begin{array}{*{35}l} x={\pi }/{6}\;+2n\pi \\ x={5\pi }/{6}\;+2n\pi \\ x={3\pi }/{2}\;+2n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer)