Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 10:07, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.4:6c moved to Solution 4.4:6c: Robot: moved page) ← Previous diff		Revision as of 11:54, 1 October 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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-	{{NAVCONTENT_START}}	+	If we use the trigonometric relation
-	<center> [[Image:4_4_6c.gif]] </center>	+	<math>\text{sin }\left( -x \right)=-\text{sin }x</math>, the equation can be rewritten as
-	{{NAVCONTENT_STOP}}	+
		+
		+	<math>\sin 2x=\sin \left( -x \right)</math>
		+
		+
		+	In exercise 4.4:5a, we saw that an equality of the type
		+
		+
		+	<math>\sin u=\sin v\quad </math>
		+
		+
		+	is satisfied if
		+
		+
		+	<math>u=v+2n\pi </math>
		+	or
		+	<math>u=\pi -v+2n\pi </math>
		+
		+
		+	where
		+	<math>n\text{ }</math>
		+	is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy
		+
		+
		+	<math>2x=-x+2n\pi </math>
		+	or
		+	<math>2x=\pi -\left( -x \right)+2n\pi </math>
		+
		+
		+	i.e.
		+
		+
		+	<math>3x=2n\pi </math>
		+	or
		+	<math>x=\pi +2n\pi </math>
		+
		+
		+	The solutions to the equation are thus
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{2n\pi }{3} \\
		+	x=\pi +2n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n\text{ }</math>
		+	an arbitrary integer)

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Revision as of 11:54, 1 October 2008

If we use the trigonometric relation sin −x=−sin x , the equation can be rewritten as

sin2x=sin−x 

In exercise 4.4:5a, we saw that an equality of the type

sinu=sinv

is satisfied if

u=v+2n or u=−v+2n

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

2x=−x+2n or 2x=−−x+2n 

i.e.

3x=2n or x=+2n

The solutions to the equation are thus

x=32nx=+2n  ( n an arbitrary integer)