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

\(\cos(\frac{2}{3}x-\frac{2}{3}\pi )=0\)

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

a

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

1p

○

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

1p

○

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

1p

4p

b

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

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

b

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

c

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

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

c

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

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

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

○

Balansmethode geeft \(-4\sin(\frac{3}{5}\pi x-\frac{4}{5}\pi )=4\) dus \(\sin(\frac{3}{5}\pi x-\frac{4}{5}\pi )=-1\text{.}\)

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

\(\cos^2(2t+\frac{4}{5}\pi )=1\)

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

a

\(\cos(2t+\frac{4}{5}\pi )=1∨\cos(2t+\frac{4}{5}\pi )=-1\)

1p

○

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

1p

○

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

1p

3p

b

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

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

b

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

1p

○

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

1p

○

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

1p
"