Getal & Ruimte (13e editie) - vwo wiskunde B

'Goniometrische vergelijkingen'.

vwo wiskunde B	8.3 Goniometrische vergelijkingen
Goniometrische vergelijkingen (7)	opgave 1 Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\) 3p a \(\sin(\frac{2}{3}x+\frac{5}{6}\pi )=0\) ExacteWaarde (0) 004f - Goniometrische vergelijkingen - basis - 52ms - dynamic variables a De exacte waardencirkel geeft \(\frac{2}{3}x+\frac{5}{6}\pi =k⋅\pi \) 1p ○ \(\frac{2}{3}x=-\frac{5}{6}\pi +k⋅\pi \) \(x=-1\frac{1}{4}\pi +k⋅1\frac{1}{2}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{4}\pi ∨x=1\frac{3}{4}\pi \) 1p 4p b \(-2\sin(\frac{1}{3}\pi x+\frac{1}{3}\pi )=-1\) ExacteWaarde (1) 004g - Goniometrische vergelijkingen - basis - 1ms - dynamic variables b Balansmethode geeft \(\sin(\frac{1}{3}\pi x+\frac{1}{3}\pi )=\frac{1}{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(\frac{1}{3}\pi x+\frac{1}{3}\pi =\frac{1}{6}\pi +k⋅2\pi ∨\frac{1}{3}\pi x+\frac{1}{3}\pi =\frac{5}{6}\pi +k⋅2\pi \) 1p ○ \(\frac{1}{3}\pi x=-\frac{1}{6}\pi +k⋅2\pi ∨\frac{1}{3}\pi x=\frac{1}{2}\pi +k⋅2\pi \) \(x=-\frac{1}{2}+k⋅6∨x=1\frac{1}{2}+k⋅6\) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=5\frac{1}{2}∨x=1\frac{1}{2}\) 1p 4p c \(4\sin(1\frac{1}{2}x-\frac{3}{4}\pi )=-2\sqrt{2}\) ExacteWaarde (2) 004h - Goniometrische vergelijkingen - basis - 0ms - dynamic variables c Balansmethode geeft \(\sin(1\frac{1}{2}x-\frac{3}{4}\pi )=-\frac{1}{2}\sqrt{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(1\frac{1}{2}x-\frac{3}{4}\pi =1\frac{1}{4}\pi +k⋅2\pi ∨1\frac{1}{2}x-\frac{3}{4}\pi =1\frac{3}{4}\pi +k⋅2\pi \) 1p ○ \(1\frac{1}{2}x=2\pi +k⋅2\pi ∨1\frac{1}{2}x=2\frac{1}{2}\pi +k⋅2\pi \) \(x=1\frac{1}{3}\pi +k⋅1\frac{1}{3}\pi ∨x=1\frac{2}{3}\pi +k⋅1\frac{1}{3}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{1}{3}\pi ∨x=0∨x=1\frac{2}{3}\pi ∨x=\frac{1}{3}\pi \) 1p 4p d \(3\cos(4x-\frac{1}{2}\pi )=-1\frac{1}{2}\sqrt{3}\) ExacteWaarde (3) 006x - Goniometrische vergelijkingen - basis - 0ms - dynamic variables d Balansmethode geeft \(\cos(4x-\frac{1}{2}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\) 1p ○ De exacte waardencirkel geeft \(4x-\frac{1}{2}\pi =\frac{5}{6}\pi +k⋅2\pi ∨4x-\frac{1}{2}\pi =-\frac{5}{6}\pi +k⋅2\pi \) 1p ○ \(4x=1\frac{1}{3}\pi +k⋅2\pi ∨4x=-\frac{1}{3}\pi +k⋅2\pi \) \(x=\frac{1}{3}\pi +k⋅\frac{1}{2}\pi ∨x=-\frac{1}{12}\pi +k⋅\frac{1}{2}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{3}\pi ∨x=\frac{5}{6}\pi ∨x=1\frac{1}{3}\pi ∨x=1\frac{5}{6}\pi ∨x=\frac{5}{12}\pi ∨x=\frac{11}{12}\pi ∨x=1\frac{5}{12}\pi ∨x=1\frac{11}{12}\pi \) 1p opgave 2 Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\) 4p \(2-4\sin(3x-\frac{5}{6}\pi )=-2\) ExacteWaarde (4) 006y - Goniometrische vergelijkingen - basis - 1ms - dynamic variables ○ Balansmethode geeft \(-4\sin(3x-\frac{5}{6}\pi )=-4\) dus \(\sin(3x-\frac{5}{6}\pi )=1\text{.}\) 1p ○ De exacte waardencirkel geeft \(3x-\frac{5}{6}\pi =\frac{1}{2}\pi +k⋅2\pi \) 1p ○ \(3x=1\frac{1}{3}\pi +k⋅2\pi \) \(x=\frac{4}{9}\pi +k⋅\frac{2}{3}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{4}{9}\pi ∨x=1\frac{1}{9}\pi ∨x=1\frac{7}{9}\pi \) 1p opgave 3 Los exact op. 3p a \(\cos^2(3x-\frac{4}{5}\pi )=1\) Kwadraat 006z - Goniometrische vergelijkingen - basis - 0ms - dynamic variables a \(\cos(3x-\frac{4}{5}\pi )=1∨\cos(3x-\frac{4}{5}\pi )=-1\) 1p ○ De exacte waardencirkel geeft \(3x-\frac{4}{5}\pi =k⋅2\pi ∨3x-\frac{4}{5}\pi =\pi +k⋅2\pi \) 1p ○ \(3x=\frac{4}{5}\pi +k⋅2\pi ∨3x=1\frac{4}{5}\pi +k⋅2\pi \) \(x=\frac{4}{15}\pi +k⋅\frac{2}{3}\pi ∨x=\frac{3}{5}\pi +k⋅\frac{2}{3}\pi \) 1p 3p b \(1\frac{1}{6}\cos(1\frac{1}{2}x+\frac{3}{4}\pi )\sin(2x+\frac{3}{5}\pi )=0\) ProductIsNul 0070 - Goniometrische vergelijkingen - basis - 1ms - dynamic variables b \(\cos(1\frac{1}{2}x+\frac{3}{4}\pi )=0∨\sin(2x+\frac{3}{5}\pi )=0\) 1p ○ De exacte waardencirkel geeft \(1\frac{1}{2}x+\frac{3}{4}\pi =\frac{1}{2}\pi +k⋅\pi ∨2x+\frac{3}{5}\pi =k⋅\pi \) 1p ○ \(1\frac{1}{2}x=-\frac{1}{4}\pi +k⋅\pi ∨2x=-\frac{3}{5}\pi +k⋅\pi \) \(x=-\frac{1}{6}\pi +k⋅\frac{2}{3}\pi ∨x=-\frac{3}{10}\pi +k⋅\frac{1}{2}\pi \) 1p

vwo wiskunde B

8.3 Goniometrische vergelijkingen

Goniometrische vergelijkingen (7)

opgave 1

Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\)

\(\sin(\frac{2}{3}x+\frac{5}{6}\pi )=0\)

ExacteWaarde (0)

004f - Goniometrische vergelijkingen - basis - 52ms - dynamic variables

De exacte waardencirkel geeft
\(\frac{2}{3}x+\frac{5}{6}\pi =k⋅\pi \)

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\(\frac{2}{3}x=-\frac{5}{6}\pi +k⋅\pi \)
\(x=-1\frac{1}{4}\pi +k⋅1\frac{1}{2}\pi \)

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\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{4}\pi ∨x=1\frac{3}{4}\pi \)

\(-2\sin(\frac{1}{3}\pi x+\frac{1}{3}\pi )=-1\)

ExacteWaarde (1)

004g - Goniometrische vergelijkingen - basis - 1ms - dynamic variables

Balansmethode geeft \(\sin(\frac{1}{3}\pi x+\frac{1}{3}\pi )=\frac{1}{2}\text{.}\)

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De exacte waardencirkel geeft
\(\frac{1}{3}\pi x+\frac{1}{3}\pi =\frac{1}{6}\pi +k⋅2\pi ∨\frac{1}{3}\pi x+\frac{1}{3}\pi =\frac{5}{6}\pi +k⋅2\pi \)

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\(\frac{1}{3}\pi x=-\frac{1}{6}\pi +k⋅2\pi ∨\frac{1}{3}\pi x=\frac{1}{2}\pi +k⋅2\pi \)
\(x=-\frac{1}{2}+k⋅6∨x=1\frac{1}{2}+k⋅6\)

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\(x\) in \([0, 2\pi ]\) geeft \(x=5\frac{1}{2}∨x=1\frac{1}{2}\)

\(4\sin(1\frac{1}{2}x-\frac{3}{4}\pi )=-2\sqrt{2}\)

ExacteWaarde (2)

004h - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

Balansmethode geeft \(\sin(1\frac{1}{2}x-\frac{3}{4}\pi )=-\frac{1}{2}\sqrt{2}\text{.}\)

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De exacte waardencirkel geeft
\(1\frac{1}{2}x-\frac{3}{4}\pi =1\frac{1}{4}\pi +k⋅2\pi ∨1\frac{1}{2}x-\frac{3}{4}\pi =1\frac{3}{4}\pi +k⋅2\pi \)

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\(1\frac{1}{2}x=2\pi +k⋅2\pi ∨1\frac{1}{2}x=2\frac{1}{2}\pi +k⋅2\pi \)
\(x=1\frac{1}{3}\pi +k⋅1\frac{1}{3}\pi ∨x=1\frac{2}{3}\pi +k⋅1\frac{1}{3}\pi \)

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\(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{1}{3}\pi ∨x=0∨x=1\frac{2}{3}\pi ∨x=\frac{1}{3}\pi \)

\(3\cos(4x-\frac{1}{2}\pi )=-1\frac{1}{2}\sqrt{3}\)

ExacteWaarde (3)

006x - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

Balansmethode geeft \(\cos(4x-\frac{1}{2}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\)

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De exacte waardencirkel geeft
\(4x-\frac{1}{2}\pi =\frac{5}{6}\pi +k⋅2\pi ∨4x-\frac{1}{2}\pi =-\frac{5}{6}\pi +k⋅2\pi \)

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\(4x=1\frac{1}{3}\pi +k⋅2\pi ∨4x=-\frac{1}{3}\pi +k⋅2\pi \)
\(x=\frac{1}{3}\pi +k⋅\frac{1}{2}\pi ∨x=-\frac{1}{12}\pi +k⋅\frac{1}{2}\pi \)

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\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{3}\pi ∨x=\frac{5}{6}\pi ∨x=1\frac{1}{3}\pi ∨x=1\frac{5}{6}\pi ∨x=\frac{5}{12}\pi ∨x=\frac{11}{12}\pi ∨x=1\frac{5}{12}\pi ∨x=1\frac{11}{12}\pi \)

opgave 2

Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\)

\(2-4\sin(3x-\frac{5}{6}\pi )=-2\)

ExacteWaarde (4)

006y - Goniometrische vergelijkingen - basis - 1ms - dynamic variables

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Balansmethode geeft \(-4\sin(3x-\frac{5}{6}\pi )=-4\) dus \(\sin(3x-\frac{5}{6}\pi )=1\text{.}\)

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De exacte waardencirkel geeft
\(3x-\frac{5}{6}\pi =\frac{1}{2}\pi +k⋅2\pi \)

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\(3x=1\frac{1}{3}\pi +k⋅2\pi \)
\(x=\frac{4}{9}\pi +k⋅\frac{2}{3}\pi \)

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\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{4}{9}\pi ∨x=1\frac{1}{9}\pi ∨x=1\frac{7}{9}\pi \)

opgave 3

Los exact op.

\(\cos^2(3x-\frac{4}{5}\pi )=1\)

Kwadraat

006z - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

\(\cos(3x-\frac{4}{5}\pi )=1∨\cos(3x-\frac{4}{5}\pi )=-1\)

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De exacte waardencirkel geeft
\(3x-\frac{4}{5}\pi =k⋅2\pi ∨3x-\frac{4}{5}\pi =\pi +k⋅2\pi \)

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\(3x=\frac{4}{5}\pi +k⋅2\pi ∨3x=1\frac{4}{5}\pi +k⋅2\pi \)
\(x=\frac{4}{15}\pi +k⋅\frac{2}{3}\pi ∨x=\frac{3}{5}\pi +k⋅\frac{2}{3}\pi \)

\(1\frac{1}{6}\cos(1\frac{1}{2}x+\frac{3}{4}\pi )\sin(2x+\frac{3}{5}\pi )=0\)

ProductIsNul

0070 - Goniometrische vergelijkingen - basis - 1ms - dynamic variables

\(\cos(1\frac{1}{2}x+\frac{3}{4}\pi )=0∨\sin(2x+\frac{3}{5}\pi )=0\)

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De exacte waardencirkel geeft
\(1\frac{1}{2}x+\frac{3}{4}\pi =\frac{1}{2}\pi +k⋅\pi ∨2x+\frac{3}{5}\pi =k⋅\pi \)

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\(1\frac{1}{2}x=-\frac{1}{4}\pi +k⋅\pi ∨2x=-\frac{3}{5}\pi +k⋅\pi \)
\(x=-\frac{1}{6}\pi +k⋅\frac{2}{3}\pi ∨x=-\frac{3}{10}\pi +k⋅\frac{1}{2}\pi \)