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

\(\sin(\frac{4}{5}x+\frac{3}{5}\pi )=0\)

ExacteWaarde (0)
004f - Goniometrische vergelijkingen - basis - dynamic variables

a

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

1p

○

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

1p

○

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

1p

4p

b

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

ExacteWaarde (1)
004g - Goniometrische vergelijkingen - basis - dynamic variables

b

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

c

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

ExacteWaarde (2)
004h - Goniometrische vergelijkingen - basis - dynamic variables

c

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

d

\(-3\cos(\frac{3}{4}\pi x+\frac{1}{6}\pi )=1\frac{1}{2}\sqrt{3}\)

ExacteWaarde (3)
006x - Goniometrische vergelijkingen - basis - dynamic variables

d

Balansmethode geeft \(\cos(\frac{3}{4}\pi x+\frac{1}{6}\pi )=-\frac{1}{2}\sqrt{3}\text{.}\)

1p

○

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

1p

○

\(\frac{3}{4}\pi x=\frac{2}{3}\pi +k⋅2\pi ∨\frac{3}{4}\pi x=-\pi +k⋅2\pi \)
\(x=\frac{8}{9}+k⋅2\frac{2}{3}∨x=-1\frac{1}{3}+k⋅2\frac{2}{3}\)

1p

○

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

1p

opgave 2

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

4p

\(-3-5\sin(3q+\frac{1}{6}\pi )=-8\)

ExacteWaarde (4)
006y - Goniometrische vergelijkingen - basis - dynamic variables

○

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

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

Kwadraat
006z - Goniometrische vergelijkingen - basis - dynamic variables

a

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

1p

○

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

1p

○

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

1p

3p

b

\(\frac{7}{8}\cos(1\frac{1}{2}t+\frac{3}{4}\pi )\sin(4t-\frac{1}{3}\pi )=0\)

ProductIsNul
0070 - Goniometrische vergelijkingen - basis - dynamic variables

b

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

1p

○

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

1p

○

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

1p
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