From Förberedande kurs i matematik 1

Revision as of 12:39, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_6b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_6b-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:38, 13 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	After moving the terms over to the left-hand side, so that
-	<center> [[Bild:4_4_6b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sqrt{2}\sin x\cos x-\cos x=0</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_4_6b-2(2).gif]] </center>	+	we see that we can take out a common factor <math>\cos x</math>,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}
		+
		+	and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
		+
		+
		+	<math>\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
		+
		+	{{Displayed math||<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
		+
		+	where ''n'' 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
		+
		+	{{Displayed math||<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+
		+	<math>\sqrt{2}\sin x - 1 = 0:</math>
		+
		+	If we rearrange the equation, we obtain the basic equation as <math>\sin x = 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
		+
		+	{{Displayed math||<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+
		+	All in all, the original equation has the solutions
		+
		+	{{Displayed math||<math>\left\{\begin{align}
		+	x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
		+	x &= \frac{\pi}{2}+n\pi\,,\\[5pt]
		+	x &= \frac{3\pi}{4}+2n\pi\,,
		+	\end{align}\right.</math>}}
		+
		+	where ''n'' is an arbitrary integer.

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Current revision

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 \cos x,

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

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

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

where n 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\,,

where n is an arbitrary integer.

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

If we rearrange the equation, we obtain the basic equation as \displaystyle \sin x = 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\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,

where n is an arbitrary integer.

All in all, the original equation has the solutions

\displaystyle \left\{\begin{align} x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{\pi}{2}+n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.