From Förberedande kurs i matematik 1

Revision as of 12:41, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_7a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_7a-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:32, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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.
-	<center> [[Bild:4_4_7a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_4_7a-2(2).gif]] </center>	+	{{Displayed math||<math>2t^2 + t = 1</math>}}
-	{{NAVCONTENT_STOP}}	+
		+	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}
		+	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}</math>}}
		+
		+	and then obtain the equation
		+
		+	{{Displayed math||<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.
		+
		+	{{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>}}
		+
		+
		+	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>.
		+
		+
		+	<math>\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
		+
		+	{{Displayed math||<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.
		+
		+
		+	<math>\sin x = -1</math>:
		+
		+	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>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+
		+	All of the solution to the equation are given by
		+
		+	{{Displayed math||<math>\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.</math>}}
		+
		+	where ''n'' is an arbitrary integer.

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Current revision

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.