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

\(\cos(4x-\frac{1}{6}\pi )=0\)

ExacteWaarde (0)
004f - Goniometrische vergelijkingen - basis - 41ms - dynamic variables

a

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

1p

○

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

1p

○

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

1p

4p

b

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

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

b

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

c

\(4\sin(\frac{1}{2}x-\frac{5}{6}\pi )=-2\sqrt{2}\)

ExacteWaarde (2)
004h - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

c

Balansmethode geeft \(\sin(\frac{1}{2}x-\frac{5}{6}\pi )=-\frac{1}{2}\sqrt{2}\text{.}\)

1p

○

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

1p

○

\(\frac{1}{2}x=2\frac{1}{12}\pi +k⋅2\pi ∨\frac{1}{2}x=2\frac{7}{12}\pi +k⋅2\pi \)
\(x=4\frac{1}{6}\pi +k⋅4\pi ∨x=5\frac{1}{6}\pi +k⋅4\pi \)

1p

○

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

1p

4p

d

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

ExacteWaarde (3)
006x - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

\(4+2\sin(3q+\frac{1}{4}\pi )=6\)

ExacteWaarde (4)
006y - Goniometrische vergelijkingen - basis - 1ms - dynamic variables

○

Balansmethode geeft \(2\sin(3q+\frac{1}{4}\pi )=2\) dus \(\sin(3q+\frac{1}{4}\pi )=1\text{.}\)

1p

○

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

1p

○

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

1p

○

\(q\) in \([0, 2\pi ]\) geeft \(q=\frac{1}{12}\pi ∨q=\frac{3}{4}\pi ∨q=1\frac{5}{12}\pi \)

1p

opgave 3

Los exact op.

3p

a

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

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

a

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

1p

○

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

1p

○

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

1p

3p

b

\(\frac{5}{8}\cos(3x-\frac{1}{6}\pi )\cos(1\frac{1}{2}x+\frac{1}{3}\pi )=0\)

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

b

\(\cos(3x-\frac{1}{6}\pi )=0∨\cos(1\frac{1}{2}x+\frac{1}{3}\pi )=0\)

1p

○

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

1p

○

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

1p
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