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

\(\cos(\frac{1}{3}\pi x-\frac{3}{4}\pi )=0\)

ExacteWaarde (0)
004f - Goniometrische vergelijkingen - basis - 52ms - dynamic variables

a

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

1p

○

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

1p

○

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

1p

4p

b

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

ExacteWaarde (1)
004g - Goniometrische vergelijkingen - basis - 0ms - dynamic variables

b

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

1p

○

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

1p

○

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

1p

○

\(t\) in \([0, 2\pi ]\) geeft \(t=2\pi ∨t=0\)

1p

4p

c

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

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

c

Balansmethode geeft \(\sin(\frac{3}{5}q-\frac{1}{4}\pi )=\frac{1}{2}\sqrt{2}\text{.}\)

1p

○

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

1p

○

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

1p

○

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

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

\(4+5\sin(1\frac{1}{2}x-\frac{1}{4}\pi )=9\)

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

○

Balansmethode geeft \(5\sin(1\frac{1}{2}x-\frac{1}{4}\pi )=5\) dus \(\sin(1\frac{1}{2}x-\frac{1}{4}\pi )=1\text{.}\)

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

\(\cos^2(2t-\frac{2}{5}\pi )=1\)

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

a

\(\cos(2t-\frac{2}{5}\pi )=1∨\cos(2t-\frac{2}{5}\pi )=-1\)

1p

○

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

1p

○

\(2t=\frac{2}{5}\pi +k⋅2\pi ∨2t=1\frac{2}{5}\pi +k⋅2\pi \)
\(t=\frac{1}{5}\pi +k⋅\pi ∨t=\frac{7}{10}\pi +k⋅\pi \)

1p

3p

b

\(\frac{7}{9}\sin(1\frac{1}{2}x-\frac{5}{6}\pi )\sin(\frac{2}{3}x+\frac{2}{3}\pi )=0\)

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

b

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

1p

○

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

1p

○

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

1p
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