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 \(\cos(1\frac{1}{2}t-\frac{1}{2}\pi )=0\) ExacteWaarde (0) 004f - Goniometrische vergelijkingen - basis - 52ms - dynamic variables a De exacte waardencirkel geeft \(1\frac{1}{2}t-\frac{1}{2}\pi =\frac{1}{2}\pi +k⋅\pi \) 1p ○ \(1\frac{1}{2}t=\pi +k⋅\pi \) \(t=\frac{2}{3}\pi +k⋅\frac{2}{3}\pi \) 1p ○ \(t\) in \([0, 2\pi ]\) geeft \(t=\frac{2}{3}\pi ∨t=0∨t=1\frac{1}{3}\pi ∨t=2\pi \) 1p 4p b \(-4\cos(1\frac{1}{2}\pi t+\frac{1}{2}\pi )=-2\) ExacteWaarde (1) 004g - Goniometrische vergelijkingen - basis - 0ms - dynamic variables b Balansmethode geeft \(\cos(1\frac{1}{2}\pi t+\frac{1}{2}\pi )=\frac{1}{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(1\frac{1}{2}\pi t+\frac{1}{2}\pi =\frac{1}{3}\pi +k⋅2\pi ∨1\frac{1}{2}\pi t+\frac{1}{2}\pi =-\frac{1}{3}\pi +k⋅2\pi \) 1p ○ \(1\frac{1}{2}\pi t=-\frac{1}{6}\pi +k⋅2\pi ∨1\frac{1}{2}\pi t=-\frac{5}{6}\pi +k⋅2\pi \) \(t=-\frac{1}{9}+k⋅1\frac{1}{3}∨t=-\frac{5}{9}+k⋅1\frac{1}{3}\) 1p ○ \(t\) in \([0, 2\pi ]\) geeft \(t=1\frac{2}{9}∨t=2\frac{5}{9}∨t=3\frac{8}{9}∨t=5\frac{2}{9}∨t=\frac{7}{9}∨t=2\frac{1}{9}∨t=3\frac{4}{9}∨t=4\frac{7}{9}∨t=6\frac{1}{9}\) 1p 4p c \(5\cos(\frac{3}{5}x+\frac{1}{2}\pi )=-2\frac{1}{2}\sqrt{2}\) ExacteWaarde (2) 004h - Goniometrische vergelijkingen - basis - 0ms - dynamic variables c Balansmethode geeft \(\cos(\frac{3}{5}x+\frac{1}{2}\pi )=-\frac{1}{2}\sqrt{2}\text{.}\) 1p ○ De exacte waardencirkel geeft \(\frac{3}{5}x+\frac{1}{2}\pi =\frac{3}{4}\pi +k⋅2\pi ∨\frac{3}{5}x+\frac{1}{2}\pi =1\frac{1}{4}\pi +k⋅2\pi \) 1p ○ \(\frac{3}{5}x=\frac{1}{4}\pi +k⋅2\pi ∨\frac{3}{5}x=\frac{3}{4}\pi +k⋅2\pi \) \(x=\frac{5}{12}\pi +k⋅3\frac{1}{3}\pi ∨x=1\frac{1}{4}\pi +k⋅3\frac{1}{3}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=\frac{5}{12}\pi ∨x=1\frac{1}{4}\pi \) 1p 4p d \(-2\sin(2q+\frac{1}{3}\pi )=\sqrt{3}\) ExacteWaarde (3) 006x - Goniometrische vergelijkingen - basis - 0ms - dynamic variables d Balansmethode geeft \(\sin(2q+\frac{1}{3}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\) 1p ○ De exacte waardencirkel geeft \(2q+\frac{1}{3}\pi =-\frac{1}{3}\pi +k⋅2\pi ∨2q+\frac{1}{3}\pi =-\frac{2}{3}\pi +k⋅2\pi \) 1p ○ \(2q=-\frac{2}{3}\pi +k⋅2\pi ∨2q=-\pi +k⋅2\pi \) \(q=-\frac{1}{3}\pi +k⋅\pi ∨q=-\frac{1}{2}\pi +k⋅\pi \) 1p ○ \(q\) in \([0, 2\pi ]\) geeft \(q=\frac{2}{3}\pi ∨q=1\frac{2}{3}\pi ∨q=\frac{1}{2}\pi ∨q=1\frac{1}{2}\pi \) 1p opgave 2 Bereken zo mogelijk exact de oplossingen in \([0, 2\pi ]\text{.}\) 4p \(2-3\sin(1\frac{1}{2}x+\frac{1}{2}\pi )=-1\) ExacteWaarde (4) 006y - Goniometrische vergelijkingen - basis - 1ms - dynamic variables ○ Balansmethode geeft \(-3\sin(1\frac{1}{2}x+\frac{1}{2}\pi )=-3\) dus \(\sin(1\frac{1}{2}x+\frac{1}{2}\pi )=1\text{.}\) 1p ○ De exacte waardencirkel geeft \(1\frac{1}{2}x+\frac{1}{2}\pi =\frac{1}{2}\pi +k⋅2\pi \) 1p ○ \(1\frac{1}{2}x=k⋅2\pi \) \(x=k⋅1\frac{1}{3}\pi \) 1p ○ \(x\) in \([0, 2\pi ]\) geeft \(x=0∨x=1\frac{1}{3}\pi \) 1p opgave 3 Los exact op. 3p a \(\sin^2(3q-\frac{1}{4}\pi )=1\) Kwadraat 006z - Goniometrische vergelijkingen - basis - 0ms - dynamic variables a \(\sin(3q-\frac{1}{4}\pi )=1∨\sin(3q-\frac{1}{4}\pi )=-1\) 1p ○ De exacte waardencirkel geeft \(3q-\frac{1}{4}\pi =\frac{1}{2}\pi +k⋅2\pi ∨3q-\frac{1}{4}\pi =1\frac{1}{2}\pi +k⋅2\pi \) 1p ○ \(3q=\frac{3}{4}\pi +k⋅2\pi ∨3q=1\frac{3}{4}\pi +k⋅2\pi \) \(q=\frac{1}{4}\pi +k⋅\frac{2}{3}\pi ∨q=\frac{7}{12}\pi +k⋅\frac{2}{3}\pi \) 1p 3p b \(\frac{3}{7}\sin(4x+\frac{1}{3}\pi )\cos(3x+\frac{3}{4}\pi )=0\) ProductIsNul 0070 - Goniometrische vergelijkingen - basis - 1ms - dynamic variables b \(\sin(4x+\frac{1}{3}\pi )=0∨\cos(3x+\frac{3}{4}\pi )=0\) 1p ○ De exacte waardencirkel geeft \(4x+\frac{1}{3}\pi =k⋅\pi ∨3x+\frac{3}{4}\pi =\frac{1}{2}\pi +k⋅\pi \) 1p ○ \(4x=-\frac{1}{3}\pi +k⋅\pi ∨3x=-\frac{1}{4}\pi +k⋅\pi \) \(x=-\frac{1}{12}\pi +k⋅\frac{1}{4}\pi ∨x=-\frac{1}{12}\pi +k⋅\frac{1}{3}\pi \) 1p

havo wiskunde B

8.4 Goniometrische vergelijkingen

Goniometrische vergelijkingen (7)

opgave 1

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

\(\cos(1\frac{1}{2}t-\frac{1}{2}\pi )=0\)

ExacteWaarde (0)

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

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

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

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

\(-4\cos(1\frac{1}{2}\pi t+\frac{1}{2}\pi )=-2\)

ExacteWaarde (1)

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

Balansmethode geeft \(\cos(1\frac{1}{2}\pi t+\frac{1}{2}\pi )=\frac{1}{2}\text{.}\)

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

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\(1\frac{1}{2}\pi t=-\frac{1}{6}\pi +k⋅2\pi ∨1\frac{1}{2}\pi t=-\frac{5}{6}\pi +k⋅2\pi \)
\(t=-\frac{1}{9}+k⋅1\frac{1}{3}∨t=-\frac{5}{9}+k⋅1\frac{1}{3}\)

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\(t\) in \([0, 2\pi ]\) geeft \(t=1\frac{2}{9}∨t=2\frac{5}{9}∨t=3\frac{8}{9}∨t=5\frac{2}{9}∨t=\frac{7}{9}∨t=2\frac{1}{9}∨t=3\frac{4}{9}∨t=4\frac{7}{9}∨t=6\frac{1}{9}\)

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

ExacteWaarde (2)

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

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

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

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

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

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

ExacteWaarde (3)

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

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

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

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

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

opgave 2

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

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

ExacteWaarde (4)

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

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

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

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

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

opgave 3

Los exact op.

\(\sin^2(3q-\frac{1}{4}\pi )=1\)

Kwadraat

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

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

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

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

\(\frac{3}{7}\sin(4x+\frac{1}{3}\pi )\cos(3x+\frac{3}{4}\pi )=0\)

ProductIsNul

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

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

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

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