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

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

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

a

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{12}\pi ∨x=\frac{7}{12}\pi ∨x=1\frac{1}{12}\pi ∨x=1\frac{7}{12}\pi \)

1p

4p

b

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

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

b

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{1}{4}∨x=6\frac{1}{4}∨x=2\frac{11}{12}\)

1p

4p

c

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

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

c

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{7}{8}\pi ∨x=1\frac{7}{8}\pi ∨x=\frac{1}{8}\pi ∨x=1\frac{1}{8}\pi \)

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{3}{4}\pi ∨x=1\frac{3}{4}\pi ∨x=\frac{7}{12}\pi ∨x=1\frac{7}{12}\pi \)

1p

opgave 2

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

4p

\(-5+4\cos(4x+\frac{1}{3}\pi )=-9\)

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

○

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

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

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

a

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

1p

○

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

1p

○

\(1\frac{1}{2}x=\frac{4}{5}\pi +k⋅2\pi ∨1\frac{1}{2}x=1\frac{4}{5}\pi +k⋅2\pi \)
\(x=\frac{8}{15}\pi +k⋅1\frac{1}{3}\pi ∨x=1\frac{1}{5}\pi +k⋅1\frac{1}{3}\pi \)

1p

3p

b

\(1\frac{2}{5}\cos(2x-\frac{1}{4}\pi )\cos(\frac{4}{5}x)=0\)

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

b

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

1p

○

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

1p

○

\(2x=\frac{3}{4}\pi +k⋅\pi ∨\frac{4}{5}x=\frac{1}{2}\pi +k⋅\pi \)
\(x=\frac{3}{8}\pi +k⋅\frac{1}{2}\pi ∨x=\frac{5}{8}\pi +k⋅1\frac{1}{4}\pi \)

1p
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