Leerdoelen
•Je kunt goniometrische vergelijkingen van de vorm\sin(A)=c\sin(A)=c\sin(A)=(c\sin(A)=(c\sin(A)=(c\sin(A)=(cenexact oplossen.
•Je kunt de waardenc=-\frac{1}{2}\sqrt3c=-\frac{1}{2}\sqrt{}c=-\frac{1}{2}c=-\frac{1}{2}c=-\frac{1}{2}c=-\frac{1}{2}c=-\frac{1}{2}c=-\frac{1}{2}c=-\frac{1}{2}\surd,-\frac{1}{2}\sqrt2-\frac{1}{2}\sqrt{s2}-\frac{1}{2}\sqrt{s}-\frac{1}{2}-\frac{1}{2}\surd-\frac{1}{2}\surd2-\frac{1}{2}\surd,,,\frac{1}{2}\sqrt2\frac{1}{2}\sqrt{}\frac{1}{2}\sqrt{s}\frac{1}{2}\frac{1}{2}\surd,\frac{1}{2}\sqrt3\frac{1}{2}\sqrt{}\frac{1}{2}\sqrt2\frac{1}{2}\sqrt{}\frac{1}{2}\sqrt{s}\frac{1}{2}\frac{1}{2}\surdherkennen en gebruiken bij het oplossen van deze vergelijkingen.
•Je kunt de eerste kwartcirkel van de exacte waardencirkel herleiden en zelf schetsen.
•Je kunt de algemene oplossingsmethoden voor sinus- en cosinusvergelijkingen toepassen, inclusief het gebruik van symmetrie en de term.
•Je kunt bepalen welke oplossingen binnen een gegeven domein (bijvoorbeeld\left\lbrack0,2\pi\right\rbrack) vallen.
De exacte waardencirkel
Om de waarden van het eerste kwadrant te onthouden, kun je de volgende truc gebruiken:
•Hoeken: Loop de hoeken op van klein naar groot:,\frac16\pi\frac{1}{\placeholder{}}\pi1\pi,,\frac13\pi\frac{1}{}\pi\frac12\pi\frac{1}{\placeholder{}}\pi1\pi\pi,.
•x-coördinaat (cosinus): De x-coördinaat is in het eerste kwadrant altijd positief en wordt kleiner naarmate de hoek groter wordt (je beweegt van rechts naar links). De waarden zijn:
•Bij\frac16\pi\frac{1}{\placeholder{}}\pi1\pi(de kleinste hoek na): Je bent nog niet ver van de x-as verwijderd, dus de x-coördinaat is de grootste van de speciale waarden:\frac{1}{2}\sqrt3\frac{1}{2}\sqrt3.\frac{1}{2}\sqrt{}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\surd.½√3.\frac{1}{2}\surd3Eeneenheidscirkel,alleenheteerstekwadrantweergegeven,metdehoeken0,⅙\pi,\frac14\pi,⅓\pi,\frac12\pi endebijbehorendecoördinaten..
•Bij: Dit is het midden van het kwadrant. De x-coördinaat is dan\frac{1}{2}\sqrt2\frac{1}{2}\sqrt{}\frac{1}{2}\sqrt3\frac{1}{2}\sqrt3..
•Bij\frac13\pi\frac{1}{\placeholder{}}\pi1\pi\pi: Je bent al flink richting de y-as, dus de x-coördinaat is de kleinste van de speciale waarden:.
•y-coördinaat (sinus): De y-coördinaat is in het eerste kwadrant ook altijd positief en wordt groter naarmate de hoek groter wordt (je beweegt van onder naar boven). De waarden zijn precies het omgekeerde van de x-coördinaten:
•Bij\frac16\pi\frac{1}{\placeholder{}}\pi1\pi\pi: Je bent nog niet ver op de y-as, dus de y-coördinaat is de kleinste:.
•Bij:\frac{1}{2}\sqrt2\frac{1}{2}\sqrt{}\frac{1}{2}\sqrt31\frac{1}{}\frac{1}{2}\frac{1}{2}\surd.
•Bij\frac13\pi\frac{1}{\placeholder{}}\pi1\pi: Je bent al flink richting de y-as, dus de y-coördinaat is de grootste:\frac{1}{2}\sqrt3.
Ezelsbruggetje: Voor de hoeken\frac16\pi\frac{1}{\placeholder{}}\pi,\frac14\pi\frac{1}{\placeholder{}}\pien\frac13\pi\frac{1}{\placeholder{}}\pizijn de cosinuswaarden (x-coördinaten) een aflopende reeks\frac12\sqrt3\frac12\sqrt{}\frac12\frac12\frac12\frac12\frac{1}{\placeholder{}},\frac12\sqrt2\frac12\sqrt{}\frac12\sqrt3\frac12\sqrt{32}\frac12\sqrt3en\frac12\sqrt1=\frac12\frac12\sqrt1=\frac{1}{\placeholder{}}\frac12\sqrt1=1\frac12\sqrt1=\frac12\sqrt1\frac12\sqrt{}\frac12\sqrt3. De sinuswaarden (y-coördinaten) zijn de oplopende reeks:\frac12\sqrt1=\frac12,\frac12\sqrt2en\frac12\sqrt3.
Goniometrische vergelijkingen oplossen
Voor het exact oplossen van goniometrische vergelijkingen van de vormofvolg je een vast stappenplan:
1.Maak de sinus of cosinus vrij: Zorg ervoor dat de term\sin(\ldots)\sin(..)\sin(.)\sin()\sin(.)\sin(..)\sin(...)si(...)s(...)of\cos(\ldots)co(\ldots)c(\ldots)(\ldots)\sin(\ldots)alleen staat aan één kant van de vergelijking. Dit betekent dat je eventuele factoren ervoor weghaalt door te delen of te vermenigvuldigen.
2.Bepaal de basishoek: Gebruik de exacte waardencirkel om de eerste basishoek (vaak aangeduid als) te vinden waarvoorof. Kies hierbij de kleinste positieve hoek.
3.Pas de algemene oplossing toe: Goniometrische functies zijn periodiek. Dit betekent dat er meerdere oplossingen zijn.
•Voor\sin(A)=c:
•(de basishoek plus hele rondjes)
•(vanwege de symmetrie van de sinus rond de y-as, plus hele rondjes)
•Voor\cos(A)=c:
•(de basishoek plus hele rondjes)
•(vanwege de symmetrie van de cosinus rond de x-as, plus hele rondjes) Hierbij staatvoor een willekeurig geheel getal (\ldots,,,,,,\ldots...).
4.Maak de variabele vrij: Alseen complexere uitdrukking is (bijvoorbeeld of), werk dan de vergelijkingen verder uit om de variabele (meestal) vrij te maken.
5.Controleer het domein: Bepaal welke van de algemene oplossingen binnen het opgegeven domein vallen (bijvoorbeeld\left\lbrack0,2\pi\right\rbrack). Vul verschillende waarden voorin (beginnend bij, dan,,,, etc.) totdat de oplossingen buiten het domein vallen.
Voorbeeld 1: sinusvergelijking
Bereken exact de oplossingen in\left\lbrack0,2\pi\right\rbrackvan de vergelijking2\sin\left(\frac{1}{2}x\right)=\sqrt22\sin\left(\frac{1}{2}x\right)=\sqrt{}2\sin\left(\frac{1}{2}x\right)=2\sin\left(\frac{1}{2}x\right)=2\sin\left(\frac{1}{2}x\right)=2\sin\left(\frac{1}{2}x\right)=2\sin\left(\frac{1}{2}x\right)=\surd2\sin\left(\frac{1}{2}x\right)=\surd22\sin\left(\frac{1}{2}x=\surd2\right).
1.Maak de sinus vrij: 2\sin\left(\frac{1}{2}x\right)=\sqrt2 \sin\left(\frac{1}{2}x\right)=\frac12\sqrt2\sin\left(\frac{1}{2}x\right)=\frac{1}{\placeholder{}}\sqrt2\sin\left(\frac{1}{2}x\right)=1\sqrt2\sin\left(\frac{1}{2}x\right)=\sqrt22\sin\left(\frac{1}{2}x\right)=\sqrt22=\sin\left(\frac{1}{2}x\right)=\sqrt22\sin\left(\frac{1}{2}x\right)=\sqrt2
2.Bepaal de basishoek: We zoeken een hoekbwaarvoor\sin(b)=\frac{1}{2}\sqrt2\sin()=\frac{1}{2}\sqrt2\sin b()=\frac{1}{2}\sqrt2\sin(Y)=\frac{1}{2}\sqrt2\sin(Y)=\frac{1}{2}\sqrt{}\sin(Y)=\frac{1}{2}\sin(Y)=\frac{1}{2}\sin(Y)=\frac{1}{2}\sin(Y)=\frac{1}{2}\sin(Y)=\frac{1}{2}\surd. Volgens de exacte waardencirkel is dit voorb=\frac{1}{4}\pi=\frac{1}{4}\pi.
3.Pas de algemene oplossing toe: Het argument isA=\frac12xA=\frac{1}{\placeholder{}}xA=1xA=1=xA=1=xen de basishoek isb=\frac14\pib=\frac{1}{\placeholder{}}\pib=1\pi. We passen de algemene oplossingsmethode voor sinus toe:
•\frac{1}{2}x=\frac{1}{4}\pi+k\cdot2\pi\frac{1}{2}x=\frac{1}{4}\pi+k2\pi
•\frac{1}{2}x=\pi-\frac{1}{4}\pi+k\cdot2\pi\frac{1}{2}x=\pi-\frac{1}{4}\pi+k2\pi \frac{1}{2}x=\frac{3}{4}\pi+k\cdot2\pi\frac{1}{2}x=\frac{3}{4}\pi+k2\pi
4.Maakvrij: Vermenigvuldig beide kanten van de vergelijkingen met:
•
•
5.Controleer het domein\left\lbrack0,2\pi\right\rbrack:
•Voor:
•Als, dan. Deze valt binnen het domein.
•Als, dan. Deze valt buiten het domein.
•Voorx=1\frac{1}{2}\pi+k\cdot4\pix=1\frac{1}{2}\pi+k4\pi:
•Als, dan. Deze valt binnen het domein.
•Als, dan. Deze valt buiten het domein.
De exacte oplossingen in het domein\left\lbrack0,2\pi\right\rbrackzijnen.
Voorbeeld 2: cosinusvergelijking
Bereken exact de oplossingen in\left\lbrack0,2\pi\right\rbrackvan de vergelijking2\cos(3x-\frac{1}{2}\pi)=-\sqrt32\cos(3x-\frac{1}{2}\pi)=-\sqrt{}2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-\surd.
1.Maak de cosinus vrij: 2\cos(3x-\frac{1}{2}\pi)=-\sqrt32\cos(3x-\frac{1}{2}\pi)=-\sqrt{}2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-2\cos(3x-\frac{1}{2}\pi)=-\surd \cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\sqrt3\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\sqrt{}\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\cos(3x-\frac{1}{2}\pi)=-\frac{1}{2}\surd
2.Bepaal de basishoek: We zoeken een hoekbwaarvoor\cos(b)=-\frac{1}{2}\sqrt3\cos()=-\frac{1}{2}\sqrt3\cos(Y)=-\frac{1}{2}\sqrt3\cos(Y)=-\frac{1}{2}\sqrt{}\cos(Y)=-\frac{1}{2}\cos(Y)=-\frac{1}{2}\cos(Y)=-\frac{1}{2}\cos(Y)=-\frac{1}{2}\cos(Y)=-\frac{1}{2}\surd. Uit de exacte waardencirkel (of door te redeneren dat de cosinus negatief is in het tweede en derde kwadrant, en\frac{1}{2}\sqrt3\frac{1}{2}\sqrt{}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\surdhoort bij\frac16\pi\frac{1}{\placeholder{}}\pi1\pi\pivan de x-as) vinden web=\frac56\pi=\frac56\piY=\frac56\piY=\frac{5}{\placeholder{}}\piY=5\piY=\piY=^\piY=^{5}\piY=^{5}⁄\piY=^{5}⁄\_\piY=^{5}⁄\_{}\piY=^{5}⁄\_{}\piY=^{5}⁄\_{}\piY=^{5}⁄\_{}\piY=^{5}⁄\_{}\pi.
3.Pas de algemene oplossing toe: Het argument isA=3x-\frac12\piA=3x-\frac12A=3x-\frac{1}{\placeholder{}}A=3x-1A=3x-A=3xA=3A=A=1A=12en de basishoek isb=\frac56\pib=\frac{5}{\placeholder{}}\pib=5\pib=\pib=1\pi. We passen de algemene oplossingsmethode voor cosinus toe:
•3x-\frac{1}{2}\pi=\frac56\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac55\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{56}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{56=}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{56}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{56}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{\placeholder{}}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=5\pi+k\cdot2\pi3x-\frac{1}{2}\pi=\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^{5}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^{5}⁄\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^{5}⁄\_\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^{5}⁄\_{6}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=^{5}⁄\_{6}\pi+k2\pi
•3x-\frac{1}{2}\pi=-\frac56\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-\frac{5}{\placeholder{}}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-5\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k*2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-+k\cdot2\pi3x-\frac{1}{2}\pi=5+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{\placeholder{}}+k\cdot2\pi3x-\frac{1}{2}\pi=\frac56+k\cdot2\pi3x-\frac{1}{2}\pi=\frac{5}{\placeholder{}}+k\cdot2\pi3x-\frac{1}{2}\pi=5+k\cdot2\pi3x-\frac{1}{2}\pi=+k\cdot2\pi3x-\frac{1}{2}\pi=-+k\cdot2\pi3x-\frac{1}{2}\pi=-^+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k\cdot2\pi3x-\frac{1}{2}\pi=-^{5}⁄\_{6}\pi+k2\pi
4.Maakvrij: Tel eerst aan beide kantenop (let op:\frac{1}{2}\pi=\frac36\pi\frac{1}{2}\pi=\frac{3}{\placeholder{}}\pi\frac{1}{2}\pi=3\pi\frac{1}{2}\pi=\pi\frac{1}{2}\pi=^\pi\frac{1}{2}\pi=^{3}\pi\frac{1}{2}\pi=^{3}⁄\pi\frac{1}{2}\pi=^{3}⁄\_\pi):
•3x=\frac56\pi+\frac36\pi+k\cdot2\pi3x=\frac56\pi+\frac36\pi+k2\pi3x=\frac56\pi+\frac36\pi+k*2\pi3x=\frac56\pi+\frac{3}{\placeholder{}}\pi+k*2\pi3x=\frac56\pi+3\pi+k*2\pi3x=\frac56\pi+\pi+k*2\pi3x=\frac56\pi+^\pi+k*2\pi3x=\frac56\pi+^{3}\pi+k*2\pi3x=\frac56\pi+^{3}⁄\pi+k*2\pi3x=\frac56\pi+^{3}⁄\_\pi+k*2\pi3x=\frac56\pi+^{3}⁄\_{6}\pi+k*2\pi3x=\frac{5}{\placeholder{}}\pi+^{3}⁄\_{6}\pi+k*2\pi3x=5\pi+^{3}⁄\_{6}\pi+k*2\pi3x=\pi+^{3}⁄\_{6}\pi+k*2\pi3x=^\pi+^{3}⁄\_{6}\pi+k*2\pi3x=^{5}\pi+^{3}⁄\_{6}\pi+k*2\pi3x=^{5}⁄\pi+^{3}⁄\_{6}\pi+k*2\pi3x=^{5}⁄\_\pi+^{3}⁄\_{6}\pi+k*2\pi 3x=\frac86\pi+k\cdot2\pi3x=\frac86\pi+k2\pi3x=\frac86\pi+k\&2\pi3x=\frac86\pi+k\&\cdot2\pi3x=\frac86\pi+k\&\cdot2\pi3x=\frac86\pi+k2\pi3x=\frac86\pi+k*2\pi3x=\frac{8}{\placeholder{}}\pi+k*2\pi3x=8\pi+k*2\pi3x=\pi+k*2\pi3x=^\pi+k*2\pi3x=^{8}\pi+k*2\pi3x=^{8}⁄\pi+k*2\pi3x=^{8}⁄\_\pi+k*2\pi 3x=\frac43\pi+k\cdot2\pi3x=\frac43\pi+k2\pi3x=\frac43\pi+k*2\pi3x=\frac{4}{\placeholder{}}\pi+k*2\pi3x=4\pi+k*2\pi3x=\pi+k*2\pi3x\pi+k*2\pi3x=\pi+k*2\pi3x={}\pi+k*2\pi3x=^{}{}\pi+k*2\pi3x=^4{}\pi+k*2\pi3x=^4\_{}\pi+k*2\pi3x=^{4}⁄\_{}\pi+k*2\pi3x=^{4}⁄\_{}\pi+k*2\pi
•3x=-\frac56\pi+\frac36\pi+k\cdot2\pi3x=-\frac{5}{\placeholder{}}\pi+\frac36\pi+k\cdot2\pi3x=-5\pi+\frac36\pi+k\cdot2\pi3x=-\pi+\frac36\pi+k\cdot2\pi3x=-^\pi+\frac36\pi+k\cdot2\pi3x=-^{5}\pi+\frac36\pi+k\cdot2\pi3x=-^{5}⁄\pi+\frac36\pi+k\cdot2\pi3x=-^{5}⁄\_\pi+\frac36\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+\frac36\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+\frac{3}{\placeholder{}}\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+3\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^{3}\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^{3}⁄\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^{3}⁄\_\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^{3}⁄\_{6}\pi+k\cdot2\pi3x=-^{5}⁄\_{6}\pi+^{3}⁄\_{6}\pi+k2\pi 3x=-\frac26\pi+k\cdot2\pi3x=-\frac{2}{\placeholder{}}\pi+k\cdot2\pi3x=-2\pi+k\cdot2\pi3x=-\pi+k\cdot2\pi3x=-^\pi+k\cdot2\pi3x=-^{2}\pi+k\cdot2\pi3x=-^{2}⁄\pi+k\cdot2\pi3x=-^{2}⁄\_\pi+k\cdot2\pi3x=-^{2}⁄\_{6}\pi+k\cdot2\pi3x=-^{2}⁄\_{6}\pi+k2\pi3x=-^{2}⁄\_{6}\pi+k=2\pi3x=-^{2}⁄\_{6}\pi+k=2\pi 3x=-\frac13\pi+k\cdot2\pi3x=-\frac{1}{\placeholder{}}\pi+k\cdot2\pi3x=-1\pi+k\cdot2\pi3x=-\pi+k\cdot2\pi3x=-⅓\pi+k\cdot2\pi3x=-⅓\pi+k2\pi
Deel nu beide kanten van de vergelijkingen door:
•x=\frac49\pi+k\cdot\frac23\pi x=-\frac19\pi+k\cdot\frac23\pix=-\frac19\pi+k\cdot\frac23^\pix=-\frac19\pi+k\cdot\frac23^{2}\pix=-\frac19\pi+k\cdot\frac23^{2}⁄\pix=-\frac19\pi+k\cdot\frac23^{2}⁄\_\pix=-\frac19\pi+k\cdot\frac23^{2}⁄\_{3}\pix=-\frac19\pi+k\cdot\frac{2}{\placeholder{}}^{2}⁄\_{3}\pix=-\frac19\pi+k\cdot2^{2}⁄\_{3}\pix=-\frac19\pi+k\cdot^{2}⁄\_{3}\pix=-\frac19\pi+k^{2}⁄\_{3}\pix=-\frac19\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}⁄\_\pi+k*^{2}⁄\_{3}\pi
5.Controleer het domein\left\lbrack0,2\pi\right\rbrack:
•Voorx=\frac49\pi+k\cdot\frac23\pix=\frac49\pi+k\cdot\frac23^\pix=\frac49\pi+k\cdot\frac23^{2}\pix=\frac49\pi+k\cdot\frac23^{2}⁄\pix=\frac49\pi+k\cdot\frac23^{2}⁄\_\pix=\frac49\pi+k\cdot\frac23^{2}⁄\_{3}\pix=\frac49\pi+k\cdot\frac{2}{\placeholder{}}^{2}⁄\_{3}\pix=\frac49\pi+k\cdot2^{2}⁄\_{3}\pix=\frac49\pi+k\cdot^{2}⁄\_{3}\pix=\frac49\pi+k^{2}⁄\_{3}\pix=\frac49\pi+k*^{2}⁄\_{3}\pix=\frac{4}{\placeholder{}}\pi+k*^{2}⁄\_{3}\pix=4\pi+k*^{2}⁄\_{3}\pix=\pi+k*^{2}⁄\_{3}\pix=^\pi+k*^{2}⁄\_{3}\pix=^{4}\pi+k*^{2}⁄\_{3}\pix=^{4}⁄\pi+k*^{2}⁄\_{3}\pix=^{4}⁄\_\pi+k*^{2}⁄\_{3}\pi(let op:\frac23{}\pi=\frac69\pi\frac23{}\pi=\frac{6}{\placeholder{}}\pi\frac23{}\pi=6\pi\frac23{}\pi=\pi\frac23{}\pi=^\pi\frac23{}\pi=^{6}\pi\frac23{}\pi=^{6}⁄\pi\frac23{}\pi=^{6}⁄\_\pi\frac23{}\pi=^{6}⁄\_{9}\pi\frac{2}{\placeholder{}}{}\pi=^{6}⁄\_{9}\pi2{}\pi=^{6}⁄\_{9}\pi{}\pi=^{6}⁄\_{9}\pi):
•k=0:x=\frac49\pik=0:x=\frac{4}{\placeholder{}}\pik=0:x=4\pik=0:x=\pik=0:x=^\pik=0:x=^{4}\pik=0:x=^{4}⁄\pik=0:x=^{4}⁄\_\pi
•k=1:x=\frac49\pi+\frac69\pi=\frac{10}{9}\pik=1:x=\frac49\pi+\frac69\pi=\frac{10}{\placeholder{}}\pik=1:x=\frac49\pi+\frac69\pi=10\pik=1:x=\frac49\pi+\frac69\pi=1\pik=1:x=\frac49\pi+\frac69\pi=\pik=1:x=\frac49\pi+\frac69\pi=^\pik=1:x=\frac49\pi+\frac69\pi=^{1}\pik=1:x=\frac49\pi+\frac69\pi=^{1}^\pik=1:x=\frac49\pi+\frac69\pi=^{1}^{0}\pik=1:x=\frac49\pi+\frac69\pi=^{1}^{0}⁄\pik=1:x=\frac49\pi+\frac69\pi=^{1}^{0}⁄\_\pik=1:x=\frac49\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=\frac{4}{\placeholder{}}\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=4\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+\frac69\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+\frac{6}{\placeholder{}}\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+6\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+^\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+^{6}\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+^{6}⁄\pi=^{1}^{0}⁄\_{9}\pik=1:x=^{4}⁄\_{9}\pi+^{6}⁄\_\pi=^{1}^{0}⁄\_{9}\pi
•k=2:x=\frac49\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=\frac{4}{\placeholder{}}\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=4\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^{4}\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+\frac{12}{9}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+\frac{12}{\placeholder{}}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+12\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+1\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^1\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=\frac{16}{9}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=\frac{16}{\placeholder{}}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=16\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=1\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{6}\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{6}⁄\pik=2:x=^{4}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{6}⁄\_\pi
•k=3:x=\frac49\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=\frac{4}{\placeholder{}}\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=4\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^{}\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\_\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+\frac{18}{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+\frac{18}{\placeholder{}}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+18\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+1\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^1\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{18}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=\frac{22}{9}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=\frac{22}{\placeholder{}}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=22\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=2\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=^{}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=^2\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=^{22}\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=^{22}⁄\pik=3:x=^4⁄\_{9}\pi+^{18}⁄\_{9}\pi=^{22}⁄\_\pik=3:x=^{4}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{2}^{2}⁄\_{}\pi(Deze waarde valt buiten het domein, omdat\frac{22}{9}>2\frac{22}{\placeholder{}}>222>2.)
•Voorx=-\frac19\pi+k\cdot\frac23\pix=-\frac19\pi\pi+k\cdot\frac23\pix=-\pi+k\cdot\frac23\pix=-^\pi+k\cdot\frac23\pix=-^{1}\pi+k\cdot\frac23\pix=-^{1}⁄\pi+k\cdot\frac23\pix=-^{1}⁄\_\pi+k\cdot\frac23\pix=-^{1}⁄\_{9}\pi+k\cdot\frac23\pix=-^{1}⁄\_{9}\pi+k\cdot *\frac23\pix=-^{1}⁄\_{9}\pi+k*\frac23\pix=-^{1}⁄\_{9}\pi+k*\frac{}{3}\pix=-^{1}⁄\_{9}\pi+k*\frac13\pix=-^{1}⁄\_{9}\pi+k*\frac{1}{}\pix=-^{1}⁄\_{9}\pi+k*\frac19\pix=-^{1}⁄\_{9}\pi+k*x=-^{1}⁄\_{9}\pi+k*^x=-^{1}⁄\_{9}\pi+k*^{2}x=-^{1}⁄\_{9}\pi+k*^{2}⁄x=-^{1}⁄\_{9}\pi+k*^{2}⁄\_x=-^{1}⁄\_{9}\pi+k*^{2}⁄\_{3}x = -¹⁄₉π + k * ²⁄₃πx=-^{1}⁄\_{9}\pi+kx=-^{1}⁄\_{9}\pi+k*x=-^{1}⁄\_{9}\pi+k*^x=-^{1}⁄\_{9}\pi+k*^{2}x=-^{1}⁄\_{9}\pi+k*^{2}⁄x=-^{1}⁄\_{9}\pi+k*^{2}⁄\_x=-^{1}⁄\_{9}\pi+k*^{2}⁄\_{3}:
•k=0:x=-\frac19\pik=0:x=-k=0:x=-^k=0:x=-^{1}k=0:x=-^{1}⁄k=0:x=-^{1}⁄\_k=0:x=-^{1}⁄\_{9}(dit is negatief, dus buiten domein)
•k=1:x=-\frac19\pi+\frac69\pi=\frac59\pik=1:x=-+\frac69\pi=\frac59\pik=1:x=-^+\frac69\pi=\frac59\pik=1:x=-^{1}+\frac69\pi=\frac59\pik=1:x=-^{1}⁄+\frac69\pi=\frac59\pik=1:x=-^{1}⁄\_+\frac69\pi=\frac59\pik=1:x=-^{1}⁄\_{9}+\frac69\pi=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+\frac69\pi=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+\frac{}{9}\pi=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+\frac19\pi=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=\frac59\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=\frac{}{9}\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=\frac19\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=k=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=^k=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=^{5}k=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=^{5}⁄k=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=^{5}⁄\pik=1:x=-^{1}⁄\_{9}\pi+^{6}⁄\_{9}\pi=^{5}⁄\_\pi
•k=2:x=-\frac19\pi+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+\frac{12}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+\frac19\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+\frac{11}{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+\frac19\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=\frac{11}{9}\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=\frac19\pik=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{1}k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{1}⁄k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{1}⁄\_k=2:x=-^{1}⁄\_{9}\pi+^{1}^{2}⁄\_{9}\pi=^{1}^{1}⁄\_{9}
•k=3:x=-\frac19\pi+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+\frac{18}{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+\frac19\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=\frac{17}{9}\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=\frac19\pik=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}^k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}^{7}k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}^{7}⁄k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}^{7}⁄\_k=3:x=-^{1}⁄\_{9}\pi+^{1}^{8}⁄\_{9}\pi=^{1}^{7}⁄\_{9}
•k=4:x=-\frac19\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-\frac{1}{}\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-\frac10\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-\frac{1}{\placeholder{}}\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-1\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+\frac{24}{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+\frac{24}{\placeholder{}}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+24\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+2\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+k=4:x=-^{1}⁄\_{9}\pi+\pi=\frac{23}{9}{}\pi\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_{9}\pi=\frac{23}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_{9}\pi=\frac29{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_{9}\pi=\frac{}{9}{}\pik=4:x=-^{1}⁄\_{9}\pi+^{2}^{4}⁄\_{9}\pi=\frac49{}\pi(Deze waarde valt buiten het domein, omdat\frac{23}{9}>2\frac{23}{\placeholder{}}>223>22>2.)
De exacte oplossingen in het domein\left\lbrack0,2\pi\right\rbrackzijnx=\frac49{}\pix=\frac49{}\pi\frac49{}\pix=\frac49{}\pix=\frac{}{9}{}\pix=\frac19{}\pix=\frac{11}{9}{}\pi,x=\frac59{}\pix=\frac{}{9}{}\pix=\frac19{}\pix=\frac{11}{9}{}\pi,x=\frac{10}{9}{}\pix=\frac19{}\pix=\frac{11}{9}{}\pi,x=\frac{11}{9}{}\pix=\frac{11}{\placeholder{}}{}\pix=11{}\pix=1{}\pix={}\pix{}\pix={}\pix=^{}{}\pix=^1{}\pix=^{11}{}\pix=^{11}⁄{}\pix=^{1}^{1}⁄\_{}\pi,x=\frac{16}{9}\pix=\frac{16}{\placeholder{}}\pix=16\pix=1\pix=\pix=^{}\pix=^1\pix=^{16}\pix=^{16}⁄\pix=^{16}⁄\_\pix=^{1}^{6}⁄\_{}\pienx=\frac{17}{9}\pix=\frac{17}{\placeholder{}}\pix=17\pix=1\pix=18\pix=1\pix=\pix\pix=\pix={}\pix=^{}{}\pix=^1{}\pix=^{17}{}\pix=^{17}⁄{}\pix=^{1}^{7}⁄\_{}\pi.
