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

\(\cos(3 x + \frac{1}{6} \pi ) = 0\)

ExacteWaarde (0)
004f - Goniometrische vergelijkingen - basis - basis - 45ms - dynamic variables

a

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

1p

○

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

1p

○

\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{9} \pi ∨ x = \frac{4}{9} \pi ∨ x = \frac{7}{9} \pi ∨ x = 1\frac{1}{9} \pi ∨ x = 1\frac{4}{9} \pi ∨ x = 1\frac{7}{9} \pi \)

1p

4p

b

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

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

b

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{5}{12} \pi ∨ x = \frac{11}{12} \pi ∨ x = 1\frac{5}{12} \pi ∨ x = 1\frac{11}{12} \pi ∨ x = \frac{1}{4} \pi ∨ x = \frac{3}{4} \pi ∨ x = 1\frac{1}{4} \pi ∨ x = 1\frac{3}{4} \pi \)

1p

4p

c

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

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

c

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

1p

○

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

1p

○

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

1p

○

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

1p

4p

d

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

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

d

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

1p

○

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

1p

○

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

1p

○

\(x\) in \([0 , 2 \pi ]\) geeft \(x = \pi ∨ x = 2 \pi \)

1p

opgave 2

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

4p

\(-3 - 4 \sin(\frac{3}{5} \pi x - \frac{1}{6} \pi ) = -7\)

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

○

(Balansmethode)
\(-4 \sin(\frac{3}{5} \pi x - \frac{1}{6} \pi ) = -4\) dus \(\sin(\frac{3}{5} \pi x - \frac{1}{6} \pi ) = 1 \text{.}\)

1p

○

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

1p

○

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

1p

○

\(x\) in \([0 , 2 \pi ]\) geeft \(x = 1\frac{1}{9} ∨ x = 4\frac{4}{9}\)

1p

opgave 3

Los exact op.

3p

a

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

Substitutie (1)
006z - Goniometrische vergelijkingen - basis - midden - 0ms - dynamic variables

a

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

1p

○

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

1p

○

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

1p

3p

b

\(\frac{2}{7} \sin(1\frac{1}{2} x - \frac{4}{5} \pi ) \sin(1\frac{1}{2} x - \frac{2}{5} \pi ) = 0\)

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

b

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

1p

○

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

1p

○

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

1p
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