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

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

ExacteWaarde (0)
004f - Goniometrische vergelijkingen - basis - basis - 72ms - dynamic variables

a

(Exacte waardencirkel)
\(4x-\frac{1}{2}\pi =k⋅\pi \)

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{8}\pi ∨x=\frac{3}{8}\pi ∨x=\frac{5}{8}\pi ∨x=\frac{7}{8}\pi ∨x=1\frac{1}{8}\pi ∨x=1\frac{3}{8}\pi ∨x=1\frac{5}{8}\pi ∨x=1\frac{7}{8}\pi \)

1p

4p

b

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

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

b

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

1p

○

(Exacte waardencirkel)
\(1\frac{1}{2}x+\frac{5}{6}\pi =\frac{1}{6}\pi +k⋅2\pi ∨1\frac{1}{2}x+\frac{5}{6}\pi =\frac{5}{6}\pi +k⋅2\pi \)

1p

○

\(1\frac{1}{2}x=-\frac{2}{3}\pi +k⋅2\pi ∨1\frac{1}{2}x=k⋅2\pi \)
\(x=-\frac{4}{9}\pi +k⋅1\frac{1}{3}\pi ∨x=k⋅1\frac{1}{3}\pi \)

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{8}{9}\pi ∨x=0∨x=1\frac{1}{3}\pi \)

1p

4p

c

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

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

c

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{3}{8}\pi ∨x=1\frac{1}{8}\pi \)

1p

4p

d

\(3\cos(\frac{2}{5}\pi x-\frac{2}{3}\pi )=1\frac{1}{2}\sqrt{3}\)

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

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

\(-3-5\cos(2x-\frac{1}{5}\pi )=2\)

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

○

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

1p

○

(Exacte waardencirkel)
\(2x-\frac{1}{5}\pi =\pi +k⋅2\pi \)

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

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

Substitutie (1)
006z - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables

a

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

1p

○

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

1p

○

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

1p

3p

b

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

Product
0070 - Goniometrische vergelijkingen - basis - midden - 1ms - dynamic variables

b

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

1p

○

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

1p

○

\(2x=\frac{3}{4}\pi +k⋅\pi ∨2x=\frac{1}{6}\pi +k⋅\pi \)
\(x=\frac{3}{8}\pi +k⋅\frac{1}{2}\pi ∨x=\frac{1}{12}\pi +k⋅\frac{1}{2}\pi \)

1p
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