Getal & Ruimte (12e editie) - havo wiskunde B

'Goniometrische vergelijkingen'.

havo wiskunde B	8.4 Goniometrische vergelijkingen
Goniometrische vergelijkingen (7)	opgave 1 Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\) 3p a \(\sin(4t-\frac{1}{3}\pi )=0\) ExacteWaarde (0) 004f - Goniometrische vergelijkingen - basis - 41ms - dynamic variables a De exacte waardencirkel geeft \(4t-\frac{1}{3}\pi =k⋅\pi \) 1p ○ \(4t=\frac{1}{3}\pi +k⋅\pi \) \(t=\frac{1}{12}\pi +k⋅\frac{1}{4}\pi \) 1p ○ \(t\) in \([0, 2\pi ]\) geeft \(t=\frac{1}{12}\pi ∨t=\frac{1}{3}\pi ∨t=\frac{7}{12}\pi ∨t=\frac{5}{6}\pi ∨t=1\frac{1}{12}\pi ∨t=1\frac{1}{3}\pi ∨t=1\frac{7}{12}\pi ∨t=1\frac{5}{6}\pi \) 1p 4p b \(4\cos(\frac{2}{3}\pi x-\frac{3}{4}\pi )=2\) ExacteWaarde (1) 004g - Goniometrische vergelijkingen - basis - 1ms - dynamic variables b Balansmethode geeft \(\cos(\frac{2}{3}\pi x-\frac{3}{4}\pi )=\frac{1}{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(\frac{2}{3}\pi x-\frac{3}{4}\pi =\frac{1}{3}\pi +k⋅2\pi ∨\frac{2}{3}\pi x-\frac{3}{4}\pi =-\frac{1}{3}\pi +k⋅2\pi \) 1p ○ \(\frac{2}{3}\pi x=1\frac{1}{12}\pi +k⋅2\pi ∨\frac{2}{3}\pi x=\frac{5}{12}\pi +k⋅2\pi \) \(x=1\frac{5}{8}+k⋅3∨x=\frac{5}{8}+k⋅3\) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{5}{8}∨x=4\frac{5}{8}∨x=\frac{5}{8}∨x=3\frac{5}{8}\) 1p 4p c \(-5\sin(3x-\frac{3}{4}\pi )=2\frac{1}{2}\sqrt{2}\) ExacteWaarde (2) 004h - Goniometrische vergelijkingen - basis - 0ms - dynamic variables c Balansmethode geeft \(\sin(3x-\frac{3}{4}\pi )=-\frac{1}{2}\sqrt{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(3x-\frac{3}{4}\pi =1\frac{1}{4}\pi +k⋅2\pi ∨3x-\frac{3}{4}\pi =1\frac{3}{4}\pi +k⋅2\pi \) 1p ○ \(3x=2\pi +k⋅2\pi ∨3x=2\frac{1}{2}\pi +k⋅2\pi \) \(x=\frac{2}{3}\pi +k⋅\frac{2}{3}\pi ∨x=\frac{5}{6}\pi +k⋅\frac{2}{3}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{2}{3}\pi ∨x=0∨x=1\frac{1}{3}\pi ∨x=2\pi ∨x=\frac{5}{6}\pi ∨x=\frac{1}{6}\pi ∨x=1\frac{1}{2}\pi \) 1p 4p d \(-2\cos(\frac{2}{5}x+\frac{2}{3}\pi )=\sqrt{3}\) ExacteWaarde (3) 006x - Goniometrische vergelijkingen - basis - 0ms - dynamic variables d Balansmethode geeft \(\cos(\frac{2}{5}x+\frac{2}{3}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\) 1p ○ De exacte waardencirkel geeft \(\frac{2}{5}x+\frac{2}{3}\pi =\frac{5}{6}\pi +k⋅2\pi ∨\frac{2}{5}x+\frac{2}{3}\pi =-\frac{5}{6}\pi +k⋅2\pi \) 1p ○ \(\frac{2}{5}x=\frac{1}{6}\pi +k⋅2\pi ∨\frac{2}{5}x=-1\frac{1}{2}\pi +k⋅2\pi \) \(x=\frac{5}{12}\pi +k⋅5\pi ∨x=-3\frac{3}{4}\pi +k⋅5\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{5}{12}\pi ∨x=1\frac{1}{4}\pi \) 1p opgave 2 Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\) 4p \(-1+3\sin(3q-\frac{1}{4}\pi )=-4\) ExacteWaarde (4) 006y - Goniometrische vergelijkingen - basis - 1ms - dynamic variables ○ Balansmethode geeft \(3\sin(3q-\frac{1}{4}\pi )=-3\) dus \(\sin(3q-\frac{1}{4}\pi )=-1\text{.}\) 1p ○ De exacte waardencirkel geeft \(3q-\frac{1}{4}\pi =1\frac{1}{2}\pi +k⋅2\pi \) 1p ○ \(3q=1\frac{3}{4}\pi +k⋅2\pi \) \(q=\frac{7}{12}\pi +k⋅\frac{2}{3}\pi \) 1p ○ \(q\) in \([0, 2\pi ]\) geeft \(q=\frac{7}{12}\pi ∨q=1\frac{1}{4}\pi ∨q=1\frac{11}{12}\pi \) 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 ○ \(3x=-\frac{4}{5}\pi +k⋅2\pi ∨3x=\frac{1}{5}\pi +k⋅2\pi \) \(x=-\frac{4}{15}\pi +k⋅\frac{2}{3}\pi ∨x=\frac{1}{15}\pi +k⋅\frac{2}{3}\pi \) 1p 3p b \(\frac{5}{9}\cos(\frac{3}{4}t-\frac{3}{4}\pi )\sin(\frac{2}{3}t+\frac{5}{6}\pi )=0\) ProductIsNul 0070 - Goniometrische vergelijkingen - basis - 1ms - dynamic variables b \(\cos(\frac{3}{4}t-\frac{3}{4}\pi )=0∨\sin(\frac{2}{3}t+\frac{5}{6}\pi )=0\) 1p ○ De exacte waardencirkel geeft \(\frac{3}{4}t-\frac{3}{4}\pi =\frac{1}{2}\pi +k⋅\pi ∨\frac{2}{3}t+\frac{5}{6}\pi =k⋅\pi \) 1p ○ \(\frac{3}{4}t=1\frac{1}{4}\pi +k⋅\pi ∨\frac{2}{3}t=-\frac{5}{6}\pi +k⋅\pi \) \(t=1\frac{2}{3}\pi +k⋅1\frac{1}{3}\pi ∨t=-1\frac{1}{4}\pi +k⋅1\frac{1}{2}\pi \) 1p

havo wiskunde B

8.4 Goniometrische vergelijkingen

Goniometrische vergelijkingen (7)

opgave 1

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

\(\sin(4t-\frac{1}{3}\pi )=0\)

ExacteWaarde (0)

004f - Goniometrische vergelijkingen - basis - 41ms - dynamic variables

De exacte waardencirkel geeft
\(4t-\frac{1}{3}\pi =k⋅\pi \)

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\(4t=\frac{1}{3}\pi +k⋅\pi \)
\(t=\frac{1}{12}\pi +k⋅\frac{1}{4}\pi \)

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\(t\) in \([0, 2\pi ]\) geeft \(t=\frac{1}{12}\pi ∨t=\frac{1}{3}\pi ∨t=\frac{7}{12}\pi ∨t=\frac{5}{6}\pi ∨t=1\frac{1}{12}\pi ∨t=1\frac{1}{3}\pi ∨t=1\frac{7}{12}\pi ∨t=1\frac{5}{6}\pi \)

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

ExacteWaarde (1)

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

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

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

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\(\frac{2}{3}\pi x=1\frac{1}{12}\pi +k⋅2\pi ∨\frac{2}{3}\pi x=\frac{5}{12}\pi +k⋅2\pi \)
\(x=1\frac{5}{8}+k⋅3∨x=\frac{5}{8}+k⋅3\)

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\(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{5}{8}∨x=4\frac{5}{8}∨x=\frac{5}{8}∨x=3\frac{5}{8}\)

\(-5\sin(3x-\frac{3}{4}\pi )=2\frac{1}{2}\sqrt{2}\)

ExacteWaarde (2)

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

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

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

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

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

\(-2\cos(\frac{2}{5}x+\frac{2}{3}\pi )=\sqrt{3}\)

ExacteWaarde (3)

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

Balansmethode geeft \(\cos(\frac{2}{5}x+\frac{2}{3}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\)

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

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\(\frac{2}{5}x=\frac{1}{6}\pi +k⋅2\pi ∨\frac{2}{5}x=-1\frac{1}{2}\pi +k⋅2\pi \)
\(x=\frac{5}{12}\pi +k⋅5\pi ∨x=-3\frac{3}{4}\pi +k⋅5\pi \)

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\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{5}{12}\pi ∨x=1\frac{1}{4}\pi \)

opgave 2

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

\(-1+3\sin(3q-\frac{1}{4}\pi )=-4\)

ExacteWaarde (4)

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

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

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

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\(3q=1\frac{3}{4}\pi +k⋅2\pi \)
\(q=\frac{7}{12}\pi +k⋅\frac{2}{3}\pi \)

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\(q\) in \([0, 2\pi ]\) geeft \(q=\frac{7}{12}\pi ∨q=1\frac{1}{4}\pi ∨q=1\frac{11}{12}\pi \)

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

\(\frac{5}{9}\cos(\frac{3}{4}t-\frac{3}{4}\pi )\sin(\frac{2}{3}t+\frac{5}{6}\pi )=0\)

ProductIsNul

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

\(\cos(\frac{3}{4}t-\frac{3}{4}\pi )=0∨\sin(\frac{2}{3}t+\frac{5}{6}\pi )=0\)

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De exacte waardencirkel geeft
\(\frac{3}{4}t-\frac{3}{4}\pi =\frac{1}{2}\pi +k⋅\pi ∨\frac{2}{3}t+\frac{5}{6}\pi =k⋅\pi \)

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