Getal & Ruimte (13e 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

\(\cos(\frac{3}{5}q+\frac{2}{5}\pi )=0\)

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

a

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

1p

○

\(\frac{3}{5}q=\frac{1}{10}\pi +k⋅\pi \)
\(q=\frac{1}{6}\pi +k⋅1\frac{2}{3}\pi \)

1p

○

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

1p

4p

b

\(-4\cos(1\frac{1}{2}t-\frac{1}{6}\pi )=-2\)

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

b

Balansmethode geeft \(\cos(1\frac{1}{2}t-\frac{1}{6}\pi )=\frac{1}{2}\text{.}\)

1p

○

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

1p

○

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

1p

○

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

1p

4p

c

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

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

c

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

1p

○

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

1p

○

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

1p

○

\(q\) in \([0, 2\pi ]\) geeft \(q=3∨q=0∨q=4\)

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

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

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

○

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

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

Kwadraat
006z - Goniometrische vergelijkingen - basis - dynamic variables

a

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

1p

○

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

1p

○

\(1\frac{1}{2}x=\frac{2}{5}\pi +k⋅2\pi ∨1\frac{1}{2}x=1\frac{2}{5}\pi +k⋅2\pi \)
\(x=\frac{4}{15}\pi +k⋅1\frac{1}{3}\pi ∨x=\frac{14}{15}\pi +k⋅1\frac{1}{3}\pi \)

1p

3p

b

\(\frac{4}{7}\sin(\frac{2}{3}x-\frac{1}{5}\pi )\sin(\frac{4}{5}x+\frac{3}{5}\pi )=0\)

ProductIsNul
0070 - Goniometrische vergelijkingen - basis - dynamic variables

b

\(\sin(\frac{2}{3}x-\frac{1}{5}\pi )=0∨\sin(\frac{4}{5}x+\frac{3}{5}\pi )=0\)

1p

○

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

1p

○

\(\frac{2}{3}x=\frac{1}{5}\pi +k⋅\pi ∨\frac{4}{5}x=-\frac{3}{5}\pi +k⋅\pi \)
\(x=\frac{3}{10}\pi +k⋅1\frac{1}{2}\pi ∨x=-\frac{3}{4}\pi +k⋅1\frac{1}{4}\pi \)

1p
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