Processing Math: Done

From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	If we consider for a moment the equality
-	<center> [[Image:4_4_5a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:4_4_5a-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	[[Image:4_4_5_a.gif]]	+	{{Displayed math||<math>\sin u = \sin v</math>|(*)}}
		+
		+	where ''u'' has a fixed value, there are usually two angles ''v'' in the unit circle which ensure that the equality holds,
		+
		+	{{Displayed math||<math>v=u\qquad\text{and}\qquad v=\pi-u\,\textrm{.}</math>}}
		+
		+	[[Image:4_4_5_a.gif||center]]
		+
		+	(The only exception is when <math>u = \pi/2</math> or <math>u=3\pi/2</math>, in which case <math>u</math> and <math>\pi-u</math> correspond to the same direction and there is only one angle ''v'' which satisfies the equality.)
		+
		+	We obtain all the angles ''v'' which satisfy (*) by adding multiples of <math>2\pi</math>,
		+
		+	{{Displayed math||<math>v = u+2n\pi\qquad\text{and}\qquad v = \pi-u+2n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+	If we now go back to our equation
		+
		+	{{Displayed math||<math>\sin 3x = \sin x</math>}}
		+
		+	the reasoning above shows that the equation is only satisfied when
		+
		+	{{Displayed math||<math>3x = x+2n\pi\qquad\text{or}\qquad 3x = \pi-x+2n\pi\,\textrm{.}</math>}}
		+
		+	If we make ''x'' the subject of each equation, we obtain the full solution to the equation,
		+
		+	{{Displayed math||<math>\left\{\begin{align}
		+	x &= 0+n\pi\,,\\[5pt]
		+	x &= \frac{\pi}{4}+\frac{n\pi}{2}\,\textrm{.}
		+	\end{align}\right.</math>}}

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Current revision

If we consider for a moment the equality

sinu=sinv

(*)

where u has a fixed value, there are usually two angles v in the unit circle which ensure that the equality holds,

v=uandv=−u.

(The only exception is when u=2 or u=32, in which case u and −u correspond to the same direction and there is only one angle v which satisfies the equality.)

We obtain all the angles v which satisfy (*) by adding multiples of 2,

v=u+2nandv=−u+2n

where n is an arbitrary integer.

If we now go back to our equation

sin3x=sinx

the reasoning above shows that the equation is only satisfied when

3x=x+2nor3x=−x+2n.

If we make x the subject of each equation, we obtain the full solution to the equation,

xx=0+n=4+2n.