Getal & Ruimte (12e 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

\(\cos(4x+\frac{1}{6}\pi )=0\)

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

a

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{12}\pi ∨x=\frac{1}{3}\pi ∨x=\frac{7}{12}\pi ∨x=\frac{5}{6}\pi ∨x=1\frac{1}{12}\pi ∨x=1\frac{1}{3}\pi ∨x=1\frac{7}{12}\pi ∨x=1\frac{5}{6}\pi \)

1p

4p

b

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

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

b

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

1p

○

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

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

1p

4p

c

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

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

c

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

1p

○

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

1p

○

\(\frac{1}{3}\pi x=1\frac{1}{12}\pi +k⋅2\pi ∨\frac{1}{3}\pi x=1\frac{7}{12}\pi +k⋅2\pi \)
\(x=3\frac{1}{4}+k⋅6∨x=4\frac{3}{4}+k⋅6\)

1p

○

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

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

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

1p

opgave 2

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

4p

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

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

○

(Balansmethode)
\(4\sin(2x+\frac{2}{3}\pi )=4\) dus \(\sin(2x+\frac{2}{3}\pi )=1\text{.}\)

1p

○

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

1p

○

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

1p

○

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

1p

opgave 3

Los exact op.

3p

a

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

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

a

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

1p

○

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

1p

○

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

1p

3p

b

\(3\frac{1}{2}\sin(2x+\frac{5}{6}\pi )\sin(2x+\frac{1}{4}\pi )=0\)

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

b

\(\sin(2x+\frac{5}{6}\pi )=0∨\sin(2x+\frac{1}{4}\pi )=0\)

1p

○

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

1p

○

\(2x=-\frac{5}{6}\pi +k⋅\pi ∨2x=-\frac{1}{4}\pi +k⋅\pi \)
\(x=-\frac{5}{12}\pi +k⋅\frac{1}{2}\pi ∨x=-\frac{1}{8}\pi +k⋅\frac{1}{2}\pi \)

1p
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