From Förberedande kurs i matematik 1

Revision as of 10:08, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.4:8a moved to Solution 4.4:8a: Robot: moved page) ← Previous diff		Current revision (08:12, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
-	<center> [[Image:4_4_8a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:4_4_8a-2(2).gif]] </center>	+	Then, we see that we can take a factor <math>\cos x</math> out of both terms,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\cos x\,(2\sin x-\sqrt{2}) = 0</math>}}
		+
		+	and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.
		+
		+
		+	<math>\cos x = 0</math>:
		+
		+	This equation has the general solution
		+
		+	{{Displayed math||<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
		+
		+
		+	<math>2\sin x-\sqrt{2}=0</math>:
		+
		+	If we collect <math>\sin x</math> on the left-hand side, we obtain the equation
		+	<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
		+
		+	{{Displayed math||<math>\left\{\begin{align}
		+	x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
		+	x &= \frac{3\pi}{4}+2n\pi\,,
		+	\end{align}\right.</math>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+
		+	The complete solution of the equation is
		+
		+	{{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.

---

Current revision

If we use the formula for double angles, \displaystyle \sin 2x = 2\sin x\cos x, and move all the terms over to the left-hand side, the equation becomes

\displaystyle 2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}

Then, we see that we can take a factor \displaystyle \cos x out of both terms,

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

and hence divide up the equation into two cases. The equation is satisfied either if \displaystyle \cos x = 0 or if \displaystyle 2\sin x-\sqrt{2} = 0\,.

\displaystyle \cos x = 0:

This equation has the general solution

\displaystyle x = \frac{\pi}{2}+n\pi\qquad(n is an arbitrary integer).

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

If we collect \displaystyle \sin x on the left-hand side, we obtain the equation \displaystyle \sin x = 1/\!\sqrt{2}, which has the general solution

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

where n is an arbitrary integer.

The complete solution of the equation is

\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.