Getal & Ruimte (13e 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

\(\sin(\frac{3}{4}x+\frac{3}{4}\pi )=0\)

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

a

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

1p

○

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

1p

○

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

1p

4p

b

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

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

b

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{11}{12}\pi ∨x=1\frac{11}{12}\pi ∨x=\frac{7}{12}\pi ∨x=1\frac{7}{12}\pi \)

1p

4p

c

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

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

c

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

1p

○

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

1p

○

\(1\frac{1}{2}x=\frac{3}{4}\pi +k⋅2\pi ∨1\frac{1}{2}x=1\frac{1}{4}\pi +k⋅2\pi \)
\(x=\frac{1}{2}\pi +k⋅1\frac{1}{3}\pi ∨x=\frac{5}{6}\pi +k⋅1\frac{1}{3}\pi \)

1p

○

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

1p

4p

d

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

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

d

(Balansmethode)
\(\cos(\frac{1}{4}\pi x+\frac{5}{6}\pi )=\frac{1}{2}\sqrt{3}\text{.}\)

1p

○

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

1p

○

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

1p

○

\(x\) in \([0, 2\pi ]\) geeft \(x=5\frac{1}{3}∨x=4\)

1p

opgave 2

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

4p

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

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

○

(Balansmethode)
\(-2\sin(4x-\frac{5}{6}\pi )=2\) dus \(\sin(4x-\frac{5}{6}\pi )=-1\text{.}\)

1p

○

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

1p

○

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

1p

○

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

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

○

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

1p

3p

b

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

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

b

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

1p

○

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

1p

○

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

1p
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