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β’Je kunt goniometrische vergelijkingen van de vormexact oplossen wanneergelijk is aan,of.
β’Je kunt goniometrische vergelijkingen van de vorm\cos(A)=cco(A)=cc(A)=c(A)=cexact oplossen wanneergelijk is aan,of.
β’Je kunt oplossingen van goniometrische vergelijkingen binnen een gegeven domein bepalen.
Goniometrische vergelijkingen:\sin\left(A\right)=c\sin\left(A\right)=\sin\left(A\right)=C\sin\left(A=C\right)\sin\left(=C\right)\sin=Csi=Cs=C
Wanneer je te maken krijgt met een goniometrische vergelijking van de vorm\sin\left(A\right)=c, waarbijde waarden,ofaanneemt, ben je op zoek naar de hoek. De sinus van een hoek op de eenheidscirkel komt overeen met de y-coΓΆrdinaat van het punt P op die cirkel.
\sin\left(A\right)=0\sin\left(A\right)=\sin\left(A\right)=c
Alsgelijk is aan, krijg je de vergelijking\sin\left(A\right)=0. Dit betekent dat de y-coΓΆrdinaat van punt P op de eenheidscirkelis.
De y-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (het startpunt).
β’Bij een hoek vanradialen (een half rondje verder).
β’Bij,, enzovoorts.
Dit betekent dat de y-coΓΆrdinaat elke halve cirkel weeris. We kunnen dit kort noteren als. Hierbij iseen geheel getal. Dit betekent datde waarden ...,,,,... mag aannemen.
\sin\left(A\right)=1\sin\left(A\right)=\sin\left(A\right)=c
Alsgelijk is aan, krijg je de vergelijking\sin\left(A\right)=1sinus A = 1eenheidscirkel met twee gemarkeerde punten. Eén punt op (1,0) en één punt op (-1,0).. Dit betekent dat de y-coârdinaat van punt P op de eenheidscirkelis.
De y-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (een kwart rondje).
β’Daarna pas weer een heel rondje verder.
We noteren dit alsA=\frac12\pi+k\cdot2\piA=\frac12\pi+k2\piA=\frac12\pi+2\piA=\frac12\pi+K2\piA=\frac12\pi+K*2\piA=\frac{1}{\placeholder{}}\pi+K*2\piA=1\pi+K*2\piA=\pi+K*2\piA=0\pi+K*2\piA=0.\pi+K*2\pi.
Ook hier iseen geheel getal.
\sin\left(A\right)=-1\sin\left(A\right)=-\sin\left(A\right)=\sin\left(A\right)=c
Alsgelijk is aan, krijg je de vergelijking\sin\left(A\right)=-1. Dit betekent dat de y-coΓΆrdinaat van punt P op de eenheidscirkelis.
De y-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (driekwart rondje).
β’Daarna pas weer een heel rondje verder.
We noteren dit alsA=1\frac12\pi+k\cdot2\piA=\frac12\pi+k\cdot2\piA=\frac12\pi+k2\piA=\frac12\pi+2\piA=\frac12\pi+K2\piA=\frac12\pi+K*2\piA=\frac{1}{\placeholder{}}\pi+K*2\piA=1\pi+K*2\piA=\pi+K*2\piA=1\pi+K*2\piA=1.\pi+K*2\pi. Ook hier iseen geheel getal.
Goniometrische vergelijkingen:\cos\left(A\right)=cco\left(A\right)=cc\left(A\right)=c\left(A\right)=c\sin\left(A\right)=c
Wanneer je te maken krijgt met een goniometrische vergelijking van de vorm\cos\left(A\right)=cco\left(A\right)=cc\left(A\right)=c\left(A\right)=c\sin\left(A\right)=c, waarbijde waarden,ofaanneemt, ben je op zoek naar de hoek. De cosinus van een hoek op de eenheidscirkel komt overeen met de x-coΓΆrdinaat van het punt P op die cirkel.
\cos\left(A\right)=0co\left(A\right)=0cod\left(A\right)=0cods\left(A\right)=0cod\left(A\right)=0co\left(A\right)=0c\left(A\right)=0\left(A\right)=0\sin\left(A\right)=0\sin\left(A\right)=\sin\left(A\right)=c
Alsgelijk is aan, krijg je de vergelijking\cos\left(A\right)=0. Dit betekent dat de x-coΓΆrdinaat van punt P op de eenheidscirkelis.
De x-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (een kwart rondje).
β’Bijradialen (driekwart rondje).
β’Daarna weer een half rondje verder, enzovoorts.
We noteren dit alsA=\frac12\pi+k\cdot\piA=\frac12\pi+k\piA=\frac12\pi+k*\piA=\frac{1}{\placeholder{}}\pi+k*\piA=1\pi+k*\piA=\pi+k*\piA=0\pi+k*\piA=0.\pi+k*\pi. Hierbij iseen geheel getal.
\cos\left(A\right)=1\cos\left(A\right)=\cos\left(A\right)=0
Alsgelijk is aan, krijg je de vergelijking\cos\left(A\right)=1. Dit betekent dat de x-coΓΆrdinaat van punt P op de eenheidscirkelis.
De x-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (het startpunt).
β’Daarna pas weer een heel rondje verder.
We noteren dit alsA=k\cdot2\piA=k2\piA=2\piA=K2\pi. Ook hier iseen geheel getal.
\cos\left(A\right)=-1\cos\left(A\right)=1\cos\left(A\right)=\cos\left(A\right)=0
Alsgelijk is aan -1-, krijg je de vergelijking\cos\left(A\right)=-1\cos\left(A\right)=1. Dit betekent dat de x-coΓΆrdinaat van punt P op de eenheidscirkelis.
De x-coΓΆrdinaat isop de volgende hoeken:
β’Bij een hoek vanradialen (een half rondje).
β’Daarna pas weer een heel rondje verder.
We noteren dit alsA=\pi+k\cdot2\piA=\pi+k2\piA=\pi+2\piA=\pi+K2\pi. Ook hier iseen geheel getal.
Voorbeeld 1:\sin(3x+\frac12\pi)=-1\sin(3x+\frac{1}{\placeholder{}}\pi)=-1\sin(3x+1\pi)=-1\sin(3x+\pi)=-1\sin(3x+0\pi)=-1\sin(3x+0.\pi)=-1
Gegeven is de vergelijking:\sin(3x+\frac12\pi)=-1. Bereken de exacte oplossingen in het domein\left\lbrack0,2\pi\right\rbrack.
Stap 1: Bepaal de basisoplossing voor de sinus. De vergelijking is van het type\sin\left(A\right)=c\sin\left(A\right)=\sin\left(A\right)=C\sin\left(A=C\right)\sin\left(=C\right)\sin=Csi=Cs=C, waarbijA=3x+\frac12\pijA=3x+\frac12\pijA=3x+\frac{1}{\placeholder{}}\pijA=3x+1\pijA=3x+\pijA=3x+0\pijA=3x+0.\pij A = 3x + 0.5ΟA=3x+0.5A=3x+0.5\pien. We bepalen bij welke hoek de y-coΓΆrdinaat gelijk is aan.
Dit is het geval bij een hoek van1\frac12\pi\frac12\pi\frac{1}{\placeholder{}}\pi1\pi\pi1\pi1.\piradialen.
Stap 2: Stel de uitdrukking gelijk aan de basisoplossing met de algemene periode. De algemene oplossing voor\sin\left(A\right)=-1\sin\left(A=-1\right)\sin\left(=-1\right)\sin=-1si=-1s=-1isA=1\frac12\pi+k\cdot2\piA=1\frac12\pi+k2\piA=1\frac12\pi+2\piA=1\frac12\pi+K2\piA=1\frac12\pi+K*2\piA=\frac12\pi+K*2\piA=\frac{1}{\placeholder{}}\pi+K*2\piA=1\pi+K*2\piA=1.\pi+K*2\pi. Dus, we stellen de uitdrukking tussen haakjes gelijk aan deze oplossing:3x+\frac12\pi=1\frac12\pi+k\cdot2\pi3x+\frac12\pi=1\frac12\pi+k2\pi3x+\frac12\pi=1\frac12\pi+2\pi3x+\frac12\pi=1\frac12\pi+K2\pi3x+\frac12\pi=1\frac12\pi+K*2\pi3x+\frac12\pi=\frac12\pi+K*2\pi3x+\frac12\pi=\frac{1}{\placeholder{}}\pi+K*2\pi3x+\frac12\pi=1\pi+K*2\pi3x+\frac12\pi=\pi+K*2\pi3x+\frac12\pi=1\pi+K*2\pi3x+\frac12\pi=1.\pi+K*2\pi3x+\frac12\pi=1.5\pi+K*2\pi3x+\frac{1}{\placeholder{}}\pi=1.5\pi+K*2\pi3x+1\pi=1.5\pi+K*2\pi3x+\pi=1.5\pi+K*2\pi3x+0\pi=1.5\pi+K*2\pi3x+05\pi=1.5\pi+K*2\pi.
Stap 3: Maakvrij.
Trek eerst\frac12\pi\frac{1}{\placeholder{}}\pi1\pivan beide kanten af:
β’3x=1\frac12\pi-\frac12\pi+k\cdot2\pi3x=1\frac12\pi-\frac12\pi+k2\pi3x=1\frac12\pi-\frac12\pi+2\pi3x=1\frac12\pi-\frac12\pi+K2\pi3x=1\frac12\pi-\frac12\pi+K*2\pi3x=1\frac12\pi-\frac{1}{\placeholder{}}\pi+K*2\pi3x=1\frac12\pi-1\pi+K*2\pi3x=1\frac12\pi-\pi+K*2\pi3x=1\frac12\pi-0\pi+K*2\pi3x=1\frac12\pi-0.\pi+K*2\pi3x=1\frac12\pi-0.5\pi+K*2\pi3x=\frac12\pi-0.5\pi+K*2\pi3x=\frac{1}{\placeholder{}}\pi-0.5\pi+K*2\pi3x=1\pi-0.5\pi+K*2\pi3x=1.\pi-0.5\pi+K*2\pi
β’3x=\pi+k\cdot2\pi3x=\pi+k2\pi3x=\pi+2\pi3x=\pi+K2\pi
Deel vervolgens door:
β’x=\frac{\pi}{3}+\frac{k\cdot2\pi}{3}x=\frac{\pi}{3}+\frac{(k\cdot2\pi}{3}x=\frac{\pi}{3}+\frac{(k\cdot2\pi)}{3}x=\frac{\pi}{3}+\frac{(k\cdot2\pi)}{\placeholder{}}x=\frac{\pi}{3}+(k\cdot2\pi)x=\frac{\pi}{3}+(k\cdot2\pi/)x=\frac{\pi}{3}+(k\cdot2\pi/3)x=\frac{\pi}{3}+(k\cdot\pi/3)x=\frac{\pi}{3}+(k\cdot *\pi/3)x=\frac{\pi}{3}+(k\cdot *2\pi/3)x=\frac{\pi}{3}+(k*2\pi/3)x=\frac{\pi}{3}+(*2\pi/3)x=\frac{\pi}{3}+(K*2\pi/3)x=(\frac{\pi}{3}+(K*2\pi/3)x=(\frac{\pi}{3})+(K*2\pi/3)x=(\frac{\pi}{\placeholder{}})+(K*2\pi/3)x=(\pi)+(K*2\pi/3)x=(\pi/)+(K*2\pi/3)
β’x=\frac13\pi+k\cdot\frac23\pix=\frac13\pi+k\cdot\frac{2}{\placeholder{}}\pix=\frac13\pi+k\cdot2\pix=\frac13\pi+k\cdot\pix=\frac13\pi+k\cdot(\pix=\frac13\pi+k\cdot(2\pix=\frac13\pi+k\cdot(2/\pix=\frac13\pi+k\cdot(2/3\pix=\frac13\pi+k\cdot(2/3)\pix=\frac13\pi+k(2/3)\pix=\frac13\pi+(2/3)\pix=\frac13\pi+K(2/3)\pix=\frac13\pi+K*(2/3)\pix=\frac{1}{\placeholder{}}\pi+K*(2/3)\pix=1\pi+K*(2/3)\pix=\pi+K*(2/3)\pix=(\pi+K*(2/3)\pix=(1\pi+K*(2/3)\pix=(1/\pi+K*(2/3)\pix=(1/3\pi+K*(2/3)\pi
Stap 4: Bepaal de oplossingen binnen het gegeven domein\left\lbrack0,2\pi\right\rbrack0,2\pi\left\lbrack\right.0,2\pi\left\lbrack\right\rbrack0,2\pi\left\lbrack0,2\pi\right\rbrack. We vullen verschillende gehele getallen voork\frac13\frac{1}{}\frac12\frac{1}{\placeholder{}}1in:
β’k=0:x=\frac13\pi+0\cdot\frac23\pi=\frac13\pik=0:x=\frac13\pi+0\cdot\frac23\pi=\pik=0:x=\frac13\pi+0\cdot\frac23\pi=(\pik=0:x=\frac13\pi+0\cdot\frac23\pi=(1\pik=0:x=\frac13\pi+0\cdot\frac23\pi=(1/\pik=0:x=\frac13\pi+0\cdot\frac23\pi=(1/3\pik=0:x=\frac13\pi+0\cdot\frac23\pi=(1/3)\pik=0:x=\frac13\pi+0\cdot\frac{}{3}\pi=(1/3)\pik=0:x=\frac13\pi+0\cdot\frac13\pi=(1/3)\pik=0:x=\frac13\pi+0\cdot\pi=(1/3)\pik=0:x=\frac13\pi+0\pi=(1/3)\pik=0:x=\frac13\pi+0*\pi=(1/3)\pik=0:x=\frac13\pi+0*(\pi=(1/3)\pik=0:x=\frac13\pi+0*(2\pi=(1/3)\pik=0:x=\frac13\pi+0*(2/\pi=(1/3)\pik=0:x=\frac13\pi+0*(2/3\pi=(1/3)\pik=0:x=\frac13\pi+0*(2/3)\pi=(1/3)\pik=0:x=\frac{1}{\placeholder{}}\pi+0*(2/3)\pi=(1/3)\pik=0:x=1\pi+0*(2/3)\pi=(1/3)\pik=0:x=\pi+0*(2/3)\pi=(1/3)\pik=0:x=(\pi+0*(2/3)\pi=(1/3)\pik=0:x=(1\pi+0*(2/3)\pi=(1/3)\pik=0:x=(1/\pi+0*(2/3)\pi=(1/3)\pik=0:x=(1/3\pi+0*(2/3)\pi=(1/3)\pi
β’k=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=\frac33\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=\frac{}{3}\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=\frac13\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=(\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=(3\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=(3/\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=(3/3\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac23\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac{}{3}\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\frac13\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+(\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+(2\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+(2/\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+(2/3\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\frac13\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=(\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=(1\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=(1/\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=(1/3\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac23\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac{}{3}\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\frac13\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot(\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot(2\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot(2/\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot(2/3\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1\cdot(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\frac13\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=(\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=(1\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=(1/\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pik=1:x=(1/3\pi+1*(2/3)\pi=(1/3)\pi+(2/3)\pi=(3/3)\pi=\pi
β’k=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=\frac53\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=\frac{}{3}\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=\frac13\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=(\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=(5\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=(5/\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=(5/3\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac43\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac{}{3}\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\frac13\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+(\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+(4\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+(4/\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+(4/3\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\frac13\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=(\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=(1\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=(1/\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=(1/3\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23(\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23(2\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23(2/\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23(2/3\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac23(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac{}{3}(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot\frac13(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2\cdot(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\frac13\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=(\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=(1\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=(1/\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pik=2:x=(1/3\pi+2*(2/3)\pi=(1/3)\pi+(4/3)\pi=(5/3)\pi
β’k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=\frac13\pi+2\pik=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=\frac13\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(\frac13\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(1\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(1/\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(1/3\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac63\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac{}{3}\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\frac13\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+(\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+(6\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+(6/\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+(6/3\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\frac13\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=(\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=(1\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=(1/\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=(1/3\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac23\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac{}{3}\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\frac13\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot(\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot(2\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot(2/\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot(2/3\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3\cdot(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\frac13\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=(\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=(1\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=(1/\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.k=3:x=(1/3\pi+3*(2/3)\pi=(1/3)\pi+(6/3)\pi=(1/3)\pi+2\pi.. Dit is buiten het domein\left\lbrack0,2\pi\right\rbrack.
β’k=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-\frac13\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-(\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-(1\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-(1/\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-(1/3\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac23\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac{}{3}\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\frac13\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-(\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-(2\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-(2/\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-(2/3\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=\frac13\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(\frac13\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(1\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(1/\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(1/3\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi d=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac23\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac{}{3}\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac13\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac{21}{3}\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac13\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac{1}{23}\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\frac13\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\cdot\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*()\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*(2)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*(2/)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*(2d/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\frac13\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=(\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=()\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=(1)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=(1/)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=(1/3)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=d(1/3)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pik=-1:x=(1/3)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pixk=-1:x=(1/3)\pi-1*(2/3)\pi=(1/3)\pi-(2/3)\pi=-(1/3)\pi. Dit is buiten het domein\left\lbrack0,2\pi\right\rbrack.
De exacte oplossingen in het domein\left\lbrack0,2\pi\right\rbrackzijn:x=\frac13\pix=\pix=(\pix=(1\pix=(1/\pix=(1/3\pi,enx=\frac53\pix=\frac{}{3}\pix=\frac13\pix=\pix=(\pix=(5\pix=(5/\pix=(5/3\pi.
Voorbeeld 2:\cos^2{}(4x)=1\cos{}(4x)=1co{}(4x)=1c{}(4x)=1
Gegeven is de vergelijking:\cos^2{}(4x)=1. Bereken exact de oplossingen in het domein\left\lbrack0,\pi\right\rbrack.
Stap 1: Verwijder het kwadraat. Om het kwadraat te verwijderen, nemen we de wortel van beide zijden. Vergeet niet dat er twee mogelijkheden zijn:\cos(4x)=1co(4x)=1c(4x)=1of\cos(4x)=-1\cos(4x)=1.
Stap 2: Bepaal de basisoplossingen voor beide gevallen.
β’Geval 1:\cos(4x)=1 We bepalen bij welke hoek de x-coΓΆrdinaat gelijk is aan. Dit is het geval bij een hoek vanradialen, en daarna elkeradialen. Dus:4x=k\cdot2\pi4x=k2\pi4x=2\pi4x=K2\pi Deel door:x=k\cdot\frac{2\pi}{4}=k\cdot\frac12\pix=k\cdot\frac{2\pi}{4}=k\cdot\frac{1}{\placeholder{}}\pix=k\cdot\frac{2\pi}{4}=k\cdot1\pix=k\cdot\frac{2\pi}{4}=k\cdot\pix=k\cdot\frac{2\pi}{4}=k\cdot(\pix=k\cdot\frac{2\pi}{4}=k\cdot(1\pix=k\cdot\frac{2\pi}{4}=k\cdot(\pix=k\cdot\frac{2\pi}{4}=k\cdot(1\pix=k\cdot\frac{2\pi}{4}=k\cdot(1/\pix=k\cdot\frac{2\pi}{4}=k\cdot(1/2\pix=k\cdot\frac{2\pi}{4}=k\cdot(1/2)\pix=k\cdot\frac{2\pi}{4}=k(1/2)\pix=k\cdot\frac{2\pi}{4}=(1/2)\pix=k\cdot\frac{2\pi}{4}=K(1/2)\pix=k\cdot\frac{2\pi}{4}=K*(1/2)\pix=k\cdot(\frac{2\pi}{4}=K*(1/2)\pix=k\cdot(\frac{2\pi}{4})=K*(1/2)\pix=k\cdot(\frac{2\pi}{\placeholder{}})=K*(1/2)\pix=k\cdot(2\pi)=K*(1/2)\pix=k\cdot(2\pi/)=K*(1/2)\pix=k\cdot(2\pi/4)=K*(1/2)\pix=k(2\pi/4)=K*(1/2)\pix=(2\pi/4)=K*(1/2)\pix=K(2\pi/4)=K*(1/2)\pi
β’Geval 2:\cos(4x)=-1\cos(4x)=1 We bepalen bij welke hoek de x-coΓΆrdinaat gelijk is aan. Dit is het geval bij een hoek vanradialen, en daarna elkeradialen. Dus:4x=\pi+k\cdot2\pi4x=\pi+k2\pi4x=\pi+2\pi4x=\pi+K2\pi Deel door: x=\frac{\pi}{4}+k\cdot\frac{2\pi}{4}x=\frac{\pi}{4}+k\cdot\frac{2\pi}{\placeholder{}}x=\frac{\pi}{4}+k\cdot2\pix=\frac{\pi}{4}+k\cdot2\pi/x=\frac{\pi}{4}+k\cdot2\pi/4x=\frac{\pi}{4}+k\cdot2\pi/4)x=\frac{\pi}{4}+(k\cdot2\pi/4)x=\frac{\pi}{\placeholder{}}+(k\cdot2\pi/4)x=\pi+(k\cdot2\pi/4)x=\pi/+(k\cdot2\pi/4)x=\pi/4+(k\cdot2\pi/4)x=(\pi/4+(k\cdot2\pi/4)x=(\pi/4)+(k\cdot2\pi/4)x=(\pi/4)+(k\cdot *2\pi/4)x=(\pi/4)+(k*2\pi/4)x=(\pi/4)+(*2\pi/4) x=\frac14\pi+k\cdot\frac12\pix=\frac14\pi+k\cdot\frac{1}{\placeholder{}}\pix=\frac14\pi+k\cdot1\pix=\frac14\pi+k\cdot\pix=\frac14\pi+k\pix=\frac14\pi+\pix=\frac14\pi+K\pix=\frac14\pi+K*\pix=\frac14\pi+K*(\pix=\frac14\pi+K*(1\pix=\frac14\pi+K*(1/\pix=\frac14\pi+K*(1/2\pix=\frac14\pi+K*(1/2)\pix=\frac{1}{\placeholder{}}\pi+K*(1/2)\pix=1\pi+K*(1/2)\pix=\pi+K*(1/2)\pix=(\pi+K*(1/2)\pix=(1\pi+K*(1/2)\pix=(1/\pi+K*(1/2)\pix=(1/4\pi+K*(1/2)\pi
Stap 3: Combineer de oplossingen (optioneel, maar efficiënt). De oplossingen zijn,\frac14\pi\frac{1}{\placeholder{}}\pi1\pi\pi5\pi,\frac12\pi\frac{1}{\placeholder{}}\pi1\pi,\frac34\pi\frac{3}{\placeholder{}}\pi3\pi,,1\frac14\pi\frac14\pi\frac{1}{\placeholder{}}\pi1\pi, enzovoorts. Dit kan korter worden genoteerd als één algemene oplossing:x=k\cdot\frac14\pix=k\cdot\frac{1}{\placeholder{}}\pix=k\cdot1\pix=k\cdot\pix=k\pix=\pix=K\pix=K*\pix=K*(\pix=K*(1\pix=K*(1/\pix=K*(1/4\pi. Dit is niet verplicht; je kunt ook de twee losse oplossingsreeksen gebruiken.
Stap 4: Bepaal de oplossingen binnen het gegeven domein\left\lbrack0,\pi\right\rbrack. We gebruiken de gecombineerde oplossingx=k\cdot\frac14\pi:
β’k=0:x=0\cdot\frac14\pi=0k=0:x=0\cdot\pi=0k=0:x=0\pi=0k=0:x=0*\pi=0k=0:x=0*(\pi=0k=0:x=0*(1\pi=0k=0:x=0*(1/\pi=0k=0:x=0*(1/4\pi=0
β’k=1:x=1\cdot\frac14\pi=\frac14\pik=1:x=1\cdot\frac14\pi=\frac14k=1:x=1\cdot\frac14\pi=k=1:x=1\cdot\frac14\pik=1:x=1\cdot\frac14(\pik=1:x=1\cdot\frac14(1\pik=1:x=1\cdot\frac14(1/\pik=1:x=1\cdot\frac14(1/4\pik=1:x=1\cdot\frac14(1/4)\pik=1:x=1\cdot\frac14(1/4)(1/4)\pik=1:x=1\cdot(1/4)(1/4)\pik=1:x=1(1/4)(1/4)\pik=1:x=1*(1/4)(1/4)\pik=1:x=1*(1/4)=(1/4)\pi
β’k=2:x=2\cdot\frac14\pi=\frac12\pik=2:x=2\cdot\frac14\pi\pi=\frac12\pik=2:x=2\cdot\frac14\pi=\pi=\frac12\pik=2:x=2\cdot\frac14\pi=3\pi=\frac12\pik=2:x=2\cdot\frac14\pi=\frac{3}{}\pi=\frac12\pik=2:x=2\cdot\frac14\pi=\frac34\pi=\frac12\pik=2:x=2\cdot\pi=\frac34\pi=\frac12\pik=2x=2\cdot\pi=\frac34\pi=\frac12\pik=x=2\cdot\pi=\frac34\pi=\frac12\pikx=2\cdot\pi=\frac34\pi=\frac12\pix=2\cdot\pi=\frac34\pi=\frac12\pix=2\cdot\pi=\frac34\pi=\frac{1}{}\pix=2\cdot\pi=\frac34\pi=\frac14\pix=2\cdot\pi=\frac34\pi=\pix=2\cdot\pi=\frac34\pi=(\pix=2\cdot\pi=\frac34\pi=(1\pix=2\cdot\pi=\frac34\pi=(1/\pix=2\cdot\pi=\frac34\pi=(1/2\pix=2\cdot\pi=\frac34\pi=(1/2)\pix=2\cdot\pi=\frac{}{4}\pi=(1/2)\pix=2\cdot\pi=\frac14\pi=(1/2)\pix=2\cdot\pi=\pi=(1/2)\pix=2\cdot\pi=(\pi=(1/2)\pix=2\cdot\pi=(2\pi=(1/2)\pix=2\cdot\pi=(2/\pi=(1/2)\pix=2\cdot\pi=(2/4\pi=(1/2)\pix=2\cdot\pi=(2/4)\pi=(1/2)\pix=2\pi=(2/4)\pi=(1/2)\pix=2*\pi=(2/4)\pi=(1/2)\pi:x=2*\pi=(2/4)\pi=(1/2)\pik:x=2*\pi=(2/4)\pi=(1/2)\pik=:x=2*\pi=(2/4)\pi=(1/2)\pik=2:x=2*\pi=(2/4)\pi=(1/2)\pik=2:x=2*(\pi=(2/4)\pi=(1/2)\pik=2:x=2*(1\pi=(2/4)\pi=(1/2)\pik=2:x=2*(1/\pi=(2/4)\pi=(1/2)\pik=2:x=2*(1/4\pi=(2/4)\pi=(1/2)\pi
β’k=3:x=3\cdot\frac14\pi=\frac34\pik=3:x=3\cdot\frac14\pi=\frac{}{4}\pik=3:x=3\cdot\frac14\pi=\frac14\pik=3:x=3\cdot\frac14\pi=\pik=3:x=3\cdot\frac14\pi=(\pik=3:x=3\cdot\frac14\pi=(3\pik=3:x=3\cdot\frac14\pi=(3/\pik=3:x=3\cdot\frac14\pi=(3/4\pik=3:x=3\cdot\frac14\pi=(3/4)\pik=3:x=3\cdot\frac14\pi(3/4)\pik=3:x=3\cdot\pi(3/4)\pik=3:x=3\pi(3/4)\pik=3:x=3*\pi(3/4)\pik=3:x=3*(\pi(3/4)\pik=3:x=3*(1\pi(3/4)\pik=3:x=3*(1/\pi(3/4)\pik=3:x=3*(1/4\pi(3/4)\pik=3:x=3*(1/4)\pi(3/4)\pik=3:x=3*(1/4)\pi=(3/4)\pik=3:x=3*(1/4)\pi(3/4)\pi
β’k=4:x=4\cdot\frac14\pi=\frac44\pi=\pik=4:x=4\frac14\pi=\frac44\pi=\pik=4:x=4\pi=\frac44\pi=\pik=4:x=4*\pi=\frac44\pi=\pik=4:x=4*(\pi=\frac44\pi=\pik=4:x=4*(1\pi=\frac44\pi=\pik=4:x=4*(1/\pi=\frac44\pi=\pik=4:x=4*(1/4\pi=\frac44\pi=\pik=4:x=4*(1/4)\pi=\frac44\pi=\pik=4:x=4*(1/4)\pi=\frac{}{4}\pi=\pik=4:x=4*(1/4)\pi=\frac14\pi=\pik=4:x=4*(1/4)\pi=\pi=\pik=4:x=4*(1/4)\pi=(\pi=\pik=4:x=4*(1/4)\pi=(4\pi=\pik=4:x=4*(1/4)\pi=(4/\pi=\pik=4:x=4*(1/4)\pi=(4/4\pi=\pi
β’k=5:x=5\cdot\frac14\pi=\frac54\pik=5:x=5\cdot\frac14\pi=\frac{}{4}\pik=5:x=5\cdot\frac14\pi=\frac14\pik=5:x=5\cdot\frac14\pi=\pik=5:x=5\cdot\frac14\pi=(\pik=5:x=5\cdot\frac14\pi=(5\pik=5:x=5\cdot\frac14\pi=(5/\pik=5:x=5\cdot\frac14\pi=(5/4\pik=5:x=5\cdot\frac14\pi=(5/4)\pik=5:x=5\cdot\pi=(5/4)\pik=5:x=5\pi=(5/4)\pik=5:x=5*\pi=(5/4)\pik=5:x=5*(\pi=(5/4)\pik=5:x=5*(1\pi=(5/4)\pik=5:x=5*(1/\pi=(5/4)\pik=5:x=5*(1/4\pi=(5/4)\pi. Dit is buiten het domein\left\lbrack0,\pi\right\rbrack.
β’k=-1:x=-1\cdot\frac14\pi=-\frac14\pik=-1:x=-1\cdot\pi=-\frac14\pik=-1:x=-1\pi=-\frac14\pik=-1:x=-1*\pi=-\frac14\pik=-1:x=-1*(\pi=-\frac14\pik=-1:x=-1*(1\pi=-\frac14\pik=-1:x=-1*(1/\pi=-\frac14\pik=-1:x=-1*(1/4\pi=-\frac14\pik=-1:x=-1*(1/4)\pi=-\frac14\pik=-1:x=-1*(1/4)\pi=-\pik=-1:x=-1*(1/4)\pi=-(\pik=-1:x=-1*(1/4)\pi=-()\pik=-1:x=-1*(1/4)\pi=-(1)\pik=-1:x=-1*(1/4)\pi=-(1/)\pi. Dit is buiten het domein\left\lbrack0,\pi\right\rbrack.
De exacte oplossingen in het domein\left\lbrack0,\pi\right\rbrackzijn:,x=\frac14\pix=\pix=(\pix=(1\pix=(1/\pix=(1/4\pi,x=\frac12\pix=\frac{1}{}\pix=\frac14\pix=\pix=(\pix=(1\pix=(1/\pix=(1/2\pi,x=\frac34\pix=\frac{}{4}\pix=\frac14\pix=\pix=(\pix=(3\pix=(3/\pix=(3/4\pien.
