1

$2 x - 4 = \frac{1}{2} x + \sqrt{2}$ $\to$ $4 x - 8 = x + 2 \sqrt{2}$ $\to$ $3 x = 8 + 2 \sqrt{2}$ $\to$ $x = \frac{8}{3} + \frac{2}{3} \sqrt{2}$
$\frac{x}{x + 1} = ‐ 2$ $\to$ $x = ‐ 2 (x + 1)$ $\to$ $3 x = ‐ 2$ $\to$ $x = ‐ \frac{2}{3}$
$\frac{4 \frac{1}{2}}{\sqrt{x}} - 3 = 0$ $\to$ $\sqrt{x} = \frac{4 \frac{1}{2}}{3} = 1 \frac{1}{2}$ $\to$ $x = {(1 \frac{1}{2})}^{2} = \frac{9}{4} = 2 \frac{1}{4}$
$\frac{x}{4} - \frac{1}{3} = \frac{x}{5} - \frac{1}{2}$ $\to$ (alles keer $60$ ) $\to$ $15 x - 20 = 12 x - 30$ $\to$ $3 x = ‐ 10$ $\to$ $x = ‐ 3 \frac{1}{3}$

2

a

$y = 4 {(x - \frac{1}{2})}^{2} - 2$ , dus top $(\frac{1}{2}, ‐ 2)$
$y = ‐ {(x - 2)}^{2} - 3$ , dus top $(2, ‐ 3)$
$y = ‐ \frac{1}{2} {(x - 3)}^{2} + 5$ , dus top $(3,5)$
$y = \frac{1}{3} {(x + 2)}^{2} + \frac{1}{3}$ , dus top $(‐ 2, \frac{1}{3})$

b

$x^{2} + 5 x + \frac{1}{4} = 0$ $\to$ ${(x + 2 \frac{1}{2})}^{2} - 6 = 0$ $\to$ $x = ‐ 2 \frac{1}{2} - \sqrt{6}$ of $x = ‐ 2 \frac{1}{2} + \sqrt{6}$
${(x - 2 \sqrt{2})}^{2} - 8 = 1$ $\to$ ${(x - 2 \sqrt{2})}^{2} = 9$ $\to$ $x = 2 \sqrt{2} - 3$ of $x = 2 \sqrt{2} + 3$
$\frac{4}{2 x} + \frac{x^{2}}{2 x} = 3$ $\to$ $\frac{4 + x^{2}}{2 x} = 3$ $\to$ $4 + x^{2} = 6 x$ $\to$ $x^{2} - 6 x - 4 = 0$ $\to$ ${(x - 3)}^{2} = 5$ $\to$ $x = 3 - \sqrt{5}$ of $x = 3 + \sqrt{5}$
$x^{2} - 4 \sqrt{3} x + 12 = 0$ $\to$ ${(x - 2 \sqrt{3})}^{2} - 12 + 12 = 0$ $\to$ $x = 2 \sqrt{3}$

3

$x^{3} - x = 0$ $\to$ $x = 0$ of $x^{2} - 1 = 0$ $\to$ $x = ‐ 1$ of $x = 0$ of $x = 1$
$x^{3} - 2 x^{2} = x^{2} + 18 x$ $\to$ $x^{3} - 3 x^{2} - 18 x = 0$ $\to$ $x = 0$ of $x^{2} - 3 x - 18 = 0$ $\to$ $x = 0$ of $(x - 6) (x + 3) = 0$ , dus $x = ‐ 3$ of $x = 0$ of $x = 6$
$x \sqrt{x - 1} = x (x - 1)$ $\to$ $x = 0$ of $\sqrt{x - 1} = x - 1$ $\to$ $x = 0$ of $x - 1 = {(x - 1)}^{2}$ $\to$ $x = 0$ of $x - 1 = 0$ of $x - 1 = 1$ , dus $\to$ $x = 0$ of $x = 1$ of $x = 2$
$x (x - 2) (x + 2) = x + 2$ $\to$ $x + 2 = 0$ of $x (x - 2) = 1$ $\to$ $x = ‐ 2$ of $x^{2} - 2 x - 1 = 0$
$\to$ $x = ‐ 2$ of $x = 1 - \sqrt{2}$ of $x = 1 + \sqrt{2}$
$\frac{1}{x} \cdot \log (3 x) = (2 x + 1) \cdot \log (3 x)$ $\to$ $\log (3 x) = 0$ of $\frac{1}{x} = 2 x + 1$ $\to$ $3 x = 1$ of $1 = 2 x^{2} + x$ $\to$ $x = \frac{1}{3}$ of $x = \frac{1}{2}$ of $x = ‐ 1$ (voldoet niet: logaritme van negatief getal bestaat niet), dus twee oplossingen: $x = \frac{1}{3}$ of $x = \frac{1}{2}$
Kruislings vermenigvuldigen: $(x + 3) (x^{2} - 1) = (x + 3) (2 x + 2)$ $\to$ $x + 3 = 0$ of $x^{2} - 1 = 2 x + 2$ $\to$ $x = ‐ 3$ of $x^{2} - 2 x - 3 = 0$ $\to$ $x = ‐ 3$ of $x = 3$ of $x = ‐ 1$ (voldoet niet: delen door nul!), dus twee oplossingen: $x = ‐ 3$ of $x = 3$ .
(Kan ook anders: de tellers zijn nul óf de noemers zijn gelijk.)

4

a

De andere rechthoekszijde is (met Pythagoras) $\sqrt{289 - x^{2}}$ . De twee rechthoekszijden samen zijn $40 - 17 = 23$ .

b

$\sqrt{289 - x^{2}} = 23 - x$ $\to$ $289 - x^{2} = {(23 - x)}^{2} = 529 - 46 x + x^{2}$ $\to$ $x^{2} - 23 x + 120 = 0$ $\to$ $(x - 8) (x - 15) = 0$ $\to$ $x = 8$ of $x = 15$

5

a

Kwadrateren geeft: $x^{2} = x + 2 \to x^{2} - x - 2 = 0$ $\to$ $(x - 2) (x + 1) = 0$ $\to$ $x = 2$ of $x = ‐ 1$ ; alleen $x = 2$ voldoet.
Kwadrateren geeft: $x^{2} = x + 2 \to x^{2} - x - 2 = 0$ $\to$ $(x - 2) (x + 1) = 0$ $\to$ $x = 2$ of $x = ‐ 1$ ; alleen $x = ‐ 1$ voldoet.
$x = \sqrt{x} + 2 \to x - 2 = \sqrt{x}$ $\to$ $x^{2} - 4 x + 4 = x \to x^{2} - 5 x + 4 = 0$
$\to$ $x^{2} - 4 x + 4 = x \to x^{2} - 5 x + 4 = 0 \to$ $(x - 1) (x - 4) = 0 \to x = 1$ of $x = 4$ ; alleen $x = 4$ voldoet.

b

$\sqrt{x} = x - 20$ $\to$ $x = {(x - 20)}^{2} = x^{2} - 40 x + 400$ $\to$ $x^{2} - 41 x + 400 = (x - 25) (x - 16) = 0$ $\to$ $x = 25$ of $x = 16$ ; alleen $x = 25$ voldoet.
$3 \sqrt{x - 2} = x - 20$ $\to$ $9 (x - 2) = {(x - 20)}^{2} = x^{2} - 40 x + 400$ $\to$ $x^{2} - 49 x + 418 = (x - 38) (x - 11) = 0$ $\to$ $x = 38$ of $x = 11$ ; alleen $x = 38$ voldoet.

c

$5 - 4 \sqrt{1 - x} = 2 x + 1 \frac{1}{2}$ $\to$ $4 \sqrt{1 - x} = 3 \frac{1}{2} - 2 x$ $\to$ $16 (1 - x) = {(3 \frac{1}{2} - 2 x)}^{2} = 12 \frac{1}{4} - 14 x + 4 x^{2}$ $\to$ $16 x^{2} + 8 x - 15 = 0$ $\to$ $x = ‐ 1 \frac{1}{4}$ of $x = \frac{3}{4}$ ;
Snijpunten: $(‐ 1 \frac{1}{4}, ‐ 1)$ en $(\frac{3}{4},3)$

6

Gelijkheid: $x^{2} = 9$ $\to$ $x = ‐ 3$ of $x = 3$ ;
Verticale asymptoot bij $x = 0$ ; schets zelf maken;
Antwoord: $‐ 3 \leq x < 0$ of $x \geq 3$ .
Gelijkheid: $\sqrt{x + 1} = x - 5$ $\to$ $x + 1 = {(x - 5)}^{2}$ $\to$ $x^{2} - 11 x + 24 = (x - 3) (x - 8) = 0$ $\to$ $x = 3$ (voldoet niet) of $x = 8$ ;
Domein van wortelfunctie is $x \geq ‐ 1$ ; schets zelf maken;
Antwoord: $‐ 1 \leq x < 8$ .
Gelijkheid: $(x - 3) (x + 4) = x + 1$ $\to$ $x^{2} = 13$ $\to$ $x = ‐ \sqrt{13}$ of $x = \sqrt{13}$ ;
Het is een dalparabool en een lijn, dus is kleiner voor waarden tussen de snijpunten;
Antwoord: $‐ \sqrt{13} < x < \sqrt{13}$ .
Gelijkheid: $‐ 2 = ‐ x (x + 1)$ $\to$ $x^{2} + x - 2 = (x - 1) (x + 2) = 0$ $\to$ $x = ‐ 2$ of $x = 1$ ;
Verticale asymptoot bij $x = 0$ ; schets zelf maken;
Antwoord: $x \leq ‐ 2$ of $‐ 1 < x \leq 1$ .

7

a

$x + 5 = 1$ of $x + 5 = 2$ , dus $x = ‐4$ of $x = ‐3$

b

$\frac{1}{x} = 1$ of $\frac{1}{x} = 2$ , dus $x = 1$ of $x = \frac{1}{2}$

c

$x^{2} = 1$ of $x^{2} = 2$ , dus $x = 1$ , $x = ‐1$ , $x = \sqrt{2}$ of $x = ‐ \sqrt{2}$

d

$(x^{2} - 3) (x^{2} + 2) = 0$ $\to$ $x^{2} = ‐ 2$ (kan niet) of $x^{2} = 3$ , dus $x = ‐ \sqrt{3}$ of $x = \sqrt{3}$

e

$\log (x) = 1$ of $\log (x) = 2$ , dus $x = 10^{1} = 10$ of $x = 10^{2} = 100$

f

Substitutie $a = 2^{x}$ geeft $a^{2} - 3 a + 2 = 0$ , dus $2^{x} = 1$ of $2^{x} = 2$ , dus $x = 0$ of $x = 1$

8

a

$x^{4} - 4 x^{2} + 3 = 0 \to (x^{2} - 3) (x^{2} - 1) = 0$ $\to$ $x^{2} = 3$ of $x^{2} = 1$ , dus $x = ‐1$ , $x = 1$ , $x = ‐ \sqrt{3}$ of $x = \sqrt{3}$

b

$x^{6} - 7 x^{3} - 8 = 0 \to (x^{3} - 8) (x^{3} + 1) = 0$ $\to$ $x^{3} = 8$ of $x^{3} = ‐ 1$ , dus $x = 2$ of $x = ‐ 1$

c

$x^{5} - x^{3} - 2 x = 0 \to x (x^{4} - x^{2} - 2) = 0 \to x (x^{2} - 2) (x^{2} + 1) = 0$ $\to$ $x = 0$ of $x^{2} = 2$ of $x^{2} = ‐ 1$ (kan niet), dus $x = 0$ , $x = ‐ \sqrt{2}$ of $x = \sqrt{2}$

9

a

$x + \sqrt{x} = 6$ geeft $a^{2} + a = 6$ met $a = \sqrt{x}$ $\to$ $a^{2} + a - 6 = 0$ $\to$ $(a + 3) (a - 2) = 0$ $\to$ $a = ‐ 3$ of $a = 2$ , dus $\sqrt{x} = ‐ 3$ (kan niet), of $\sqrt{x} = 2$ geeft enige oplossing $x = 4$

b

Substitueer $x + 1 = a$ , je krijgt dan:
$a^{3} - a^{2} - 2 a = 0 \to a (a - 2) (a + 1) = 0$ $\to$ $a = 0$ of $a = 2$ of $a = ‐ 1$ $\to$ $x + 1 = 0$ of $x + 1 = 2$ of $x + 1 = ‐ 1$ , dus $x = ‐ 1$ , $x = 1$ of $x = ‐ 2$

c

Substitueer $a = \sqrt{x + 3}$ , dan krijg je $a - \frac{125}{a^{2}} = 0$ $\to$ $a^{3} = 125$ $\to$ $a = 5$ , dus $\sqrt{x + 3} = 5$ $\to$ $x + 3 = 25$ $\to$ $x = 22$

d

Substitueer $x + 3 = a$ , dan krijg je:
$\frac{1}{a} + a = 2 \frac{1}{2} \to 2 a^{2} - 5 a + 2 = 0 \to a = 2$ of $a = \frac{1}{2}$ , dus $x = ‐1$ of $x = ‐2 \frac{1}{2}$

e

$(\frac{1}{x} - 2) (\frac{1}{x} + 3) = 0$ $\to$ $\frac{1}{x} = 2$ of $\frac{1}{x} = ‐ 3$ , dus $x = \frac{1}{2}$ of $x = ‐ \frac{1}{3}$

f

Substitutie $a = \log (x)$ geeft $a - \frac{1}{a} = 0$ $\to$ $a^{2} - 1 = 0$ $\to$ $a = ‐ 1$ of $a = 1$ $\to$ $\log (x) = ‐ 1$ of $\log (x) = 1$ $\to$ $x = 10^{‐ 1} = 0,1$ of $x = 10^{1} = 10$

10

$x = 1$ , $y = 1$	$x = \frac{1}{2}$ , $y = ‐ \frac{1}{2}$
$x = ‐ 2$ , $y = ‐ 3$	$x = 3$ , $y = ‐ 4$

11

$f$ : punt $(0,2)$ invullen geeft direct $c = 2$ ; De twee andere punten invullen geeft het stelsel ${\begin{matrix} 16 a - 4 b + 2 = 7 \\ 16 a + 4 b + 2 = ‐ 1 \end{matrix}$ met de oplossing $a = \frac{1}{16}$ en $b = ‐ 1$ .
$g$ : punten $(0,6)$ en $(1,4)$ invullen geeft het stelsel ${\begin{matrix} \frac{p}{q} = 6 \\ \frac{p}{q + 1} = 4 \end{matrix}$ ; de bovenste vergelijking geeft $p = 6 q$ en dan substitueren in de tweede vergelijking geeft $\frac{6 q}{q + 1} = 4$ $\to$ $4 (q + 1) = 6 q$ $\to$ $q = 2$ $\to$ $p = 12$ .

12

${\begin{matrix} y = 11 - \sqrt{x} \\ y = x + 5 \end{matrix}$ $\to$ $x + 5 = 11 - \sqrt{x}$ $\to$ $\sqrt{x} = 6 - x$ $\to$ $x = 36 - 12 x + x^{2}$ $\to$ $(x - 4) (x - 9) = 0$ $\to$ $x = 9$ (voldoet niet) of $x = 4$ $\to$ $y = 9$
${\begin{matrix} y = 5 - x^{2} \\ y = {(1 - x)}^{2} \end{matrix}$ $\to$ ${(1 - x)}^{2} = 5 - x^{2}$ $\to$ $x^{2} - x - 2 = 0$ $\to$ $(x - 2) (x + 1) = 0$ $\to$ $x = 2$ of $x = ‐ 1$ . $x = 2$ geeft $y = 1$ maar dit voldoet niet;
Enige oplossing: $x = ‐ 1$ en $y = 4$ .

13

$x = 2 \frac{1}{2}$	$x = 32$
$x = 16$	$x = ‐ \frac{27}{16}$ of $x = \frac{27}{16}$
$x = 4$	$x = ‐ \frac{1}{2}$

14

a

$10^{x - 1} = 10^{1 + \frac{1}{2} x}$ $\to$ $x - 1 = 1 + \frac{1}{2} x$ $\to$ $x = 4$
$3^{\frac{1}{2} - 2 x} = 3^{‐ 1 + \frac{1}{2} x}$ $\to$ $\frac{1}{2} - 2 x = ‐ 1 + \frac{1}{2} x$ $\to$ $x = \frac{3}{5}$
$‐ \frac{2^{x}}{4} = {(‐ 2^{x})}^{3} = ‐ 2^{3 x}$ $\to$ $2^{x - 2} = 2^{3 x}$ $\to$ $x - 2 = 3 x$ $\to$ $x = ‐ 1$
$2^{\frac{1}{2} x} = 2^{\sqrt{x}}$ $\to$ $\frac{1}{2} x = \sqrt{x}$ $\to$ $x - 2 \sqrt{x} = 0$ $\to$ $\sqrt{x} (\sqrt{x} - 2) = 0$ $\to$ $x = 0$ of $x = 4$

b

Noem $2^{x} = y$ , dan $y + \frac{1}{y} = 2 \frac{1}{2}$ $\to$ $y^{2} - 2 \frac{1}{2} y + 1 = 0$ $\to$ $(y - 1) (y - \frac{1}{2}) = 0$ $\to$ $y = \frac{1}{2}$ of $y = 2$ , dus $x = 1$ of $x = ‐ 1$ .
$2^{x} + 2 \cdot 2^{‐ x} = 3$ ; noem $2^{x} = y$ , dan: $y + \frac{2}{y} = 3 \to y^{2} - 3 y + 2 = 0 \to y = 1$ of $y = 2$ , dus $x = 0$ of $x = 1$ .
$4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2}$ , dus de vergelijking wordt ${(2^{x})}^{2} + 16 = 10 \cdot 2^{x}$ ;
noem $2^{x} = y$ , dan: $y^{2} - 10 y + 16 = 0 \to y = 2$ of $y = 8$ , dus $x = 1$ of $x = 3$ .
Delen door $2^{x}$ geeft $\sqrt{3} \cdot 3^{x} = 81$ $\to$ $3^{x} = \frac{81}{\sqrt{3}} = \frac{3^{4}}{3^{\frac{1}{2}}} = 3^{3 \frac{1}{2}}$ $\to$ $x = 3 \frac{1}{2}$ .

c

${}^{5} \log (x + 1) = {}^{5} \log (25) + {}^{5} \log (x - 1) = {}^{5} \log (25 (x - 1))$ $\to$ $x + 1 = 25 (x - 1)$ $\to$ $x = \frac{13}{12}$
${}^{5} \log (x + 1) = {}^{5} \log ({(x - 1)}^{2})$ $\to$ $x + 1 = {(x - 1)}^{2} = x^{2} - 2 x + 1$ $\to$ $x (x - 3) = 0$ $\to$ $x = 0$ of $x = 3$ . Alleen $x = 3$ voldoet.
Noem $log (x) = y$ dan $\frac{1}{y} + y = 2 \frac{1}{2} \to 2 y^{2} - 5 y + 2 = 0 \to y = \frac{1}{2}$ of $y = 2$ , dus: $x = \sqrt{10}$ of $x = 100$ .
$2 \cdot log (x) - log (x + 4) = log (2 x - 6)$ $\to$ $log (x^{2}) = log ((x + 4) (2 x - 6))$ $\to$ $x^{2} = (x + 4) (2 x - 6)$ $\to$ $x^{2} + 2 x - 24 = 0 \to x = 4$ of $x = ‐ 6$ .
Alleen $x = 4$ voldoet.
$2 \cdot log (x) - log (x + 1) = log (x - 2)$ $\to$ $log (x^{2}) = log ((x + 1) (x - 2))$ $\to$ $x^{2} = (x + 1) (x - 2) \to x = ‐ 2$ .
Maar $x = ‐ 2$ voldoet niet, dus de vergelijking heeft geen oplossingen.

15

Gelijkheid: $3^{5 - 2 x} = 3^{1 \frac{1}{2}}$ $\to$ $x = 1 \frac{3}{4}$ ; schets maken (geen asymptoot) en antwoord geven: $x \leq 1 \frac{3}{4}$ .
Gelijkheid: $2^{x - 2} = 2^{\frac{2}{3}}$ $\to$ $x = 2 \frac{2}{3}$ ; schets maken (geen asymptoot) en antwoord geven: $x > 2 \frac{2}{3}$ .
Gelijkheid: $10^{2 - x} = 10^{\frac{1}{2} x}$ $\to$ $2 - x = \frac{1}{2} x$ $\to$ $x = \frac{4}{3}$ ; schets maken (geen asymptoot) en antwoord geven: $x > \frac{4}{3}$ .
Gelijkheid: $x \sqrt{x} = 64$ $\to$ $x^{\frac{3}{2}} = 2^{6}$ $\to$ $x = {(2^{6})}^{\frac{2}{3}} = 2^{4} = 16$ ; schets maken (asymptoot bij $x = 0$ ) en antwoord geven: $0 < x \leq 16$ .
Gelijkheid: $\sqrt[3]{x} = \frac{1}{2} x$ $\to$ $x = {(\frac{1}{2} x)}^{3} = \frac{1}{8} x^{3}$ $\to$ $x = 0$ of $\frac{1}{8} x^{2} = 1$ $\to$ $x = 0$ of $x = ‐ \sqrt{8}$ of $x = \sqrt{8}$ ; schets maken (geen asymptoot) en antwoord geven: $x < ‐ \sqrt{8}$ of $0 < x < \sqrt{8}$ .
Gelijkheid: $\frac{4 (x - 1) + 4 (x + 1)}{(x + 1) (x - 1)} = x$ $\to$ $\frac{8 x}{x^{2} - 1} = x$ $\to$ $x^{3} - x = 8 x$ $\to$ $x^{3} - 9 x = 0$ $\to$ $x (x^{2} - 9) = 0$ $\to$ $x = 0$ of $x = ‐ 3$ of $x = 3$ ; schets maken (twee asymptoten: bij $x = ‐ 1$ en $x = 1$ ) en antwoord geven: $x \leq ‐ 3$ of $‐ 1 < x \leq 0$ of $1 < x \leq 3$ .
Gelijkheid: $\log (x + \frac{1}{3}) = \log (10) - \log (x) = \log (\frac{10}{x})$ $\to$ $x + \frac{1}{3} = \frac{10}{x}$ $\to$ $x^{2} + \frac{1}{3} x - 10 = 0$ $\to$ $x = 3$ of $x = ‐ 3 \frac{1}{3}$ (voldoet niet); schets maken (twee verticale asymptoten bij $x = ‐ \frac{1}{3}$ en $x = 0$ ) en antwoord geven: $x \geq 3$ .
Gelijkheid: $\log (x + 2) = \log (2 - x)$ $\to$ $x + 2 = 2 - x$ $\to$ $x = 0$ ; schets maken (twee verticale asymptoten bij $x = ‐ 2$ en $x = 2$ ) en antwoord geven: $0 \leq x < 2$ .

16

a

$\sin (2 x) = ‐ \frac{1}{2}$ $\to$ $2 x = ‐ \frac{1}{6} π + k \cdot 2 π$ of $2 x = π - ‐ \frac{1}{6} π + k \cdot 2 π$ ( $k$ geheel)
$\to$ $x = ‐ \frac{1}{12} π + k \cdot π$ of $x = \frac{7}{12} π + k \cdot π$
dus $x = \frac{11}{12} π$ , $x = 1 \frac{11}{12} π$ , $x = \frac{5}{12} π$ of $x = 1 \frac{5}{12} π$
$\sin (π x) = \frac{1}{2} \sqrt{2}$ $\to$ $π x = \frac{1}{4} π + k \cdot 2 π$ of $π x = \frac{3}{4} π + k \cdot 2 π$ ( $k$ geheel)
$\to$ $x = \frac{1}{4} + 2 k$ of $x = \frac{3}{4} + 2 k$ ,
dus $x = \frac{1}{4}$ , $x = 2 \frac{1}{4}$ , $x = 4 \frac{1}{4}$ , $x = 6 \frac{1}{4}$ , $x = \frac{3}{4}$ , $x = 2 \frac{3}{4}$ of $x = 4 \frac{3}{4}$
$3 x - \frac{1}{3} π = \frac{1}{3} π + k \cdot 2 π$ of $3 x - \frac{1}{3} π = π - \frac{1}{3} π + k \cdot 2 π$ ( $k$ geheel)
$\to$ $3 x = \frac{2}{3} π + k \cdot 2 π$ of $3 x = π + k \cdot 2 π$
$\to$ $x = \frac{2}{9} π + k \cdot \frac{2}{3} π$ of $x = \frac{1}{3} π + k \cdot \frac{2}{3} π$ ,
dus $x = \frac{2}{9} π$ , $x = \frac{8}{9} π$ , $x = 1 \frac{5}{9} π$ , $x = \frac{1}{3} π$ , $x = π$ of $x = 1 \frac{2}{3} π$
$\cos (2 x) = ‐ \sqrt{\frac{3}{4}} = ‐ \frac{1}{2} \sqrt{3}$ of $\cos (2 x) = \frac{1}{2} \sqrt{3}$
$\to$ $2 x = \frac{1}{6} π + k \cdot 2 π$ of $2 x = 2 π - \frac{1}{6} π + k \cdot 2 π$ ( $k$ geheel)
$\to$ $x = \frac{1}{12} π + k \cdot π$ of $x = \frac{11}{12} π + k \cdot π$ ,
dus $x = \frac{1}{12} π$ , $x = \frac{11}{12} π$ , $x = 1 \frac{1}{12} π$ of $x = 1 \frac{11}{12} π$

b

$\sin (0,4 x - 1) = 0,4$
$\to$ $0,4 x - 1 = 0,4115... + k \cdot 2 π$ of $0,4 x - 1 = π - 0,4115... + k \cdot 2 π$ ( $k$ geheel) $\to$ $x = 3,52879... + k \cdot 5 π$ of $x = 9,325... + k \cdot 5 π$ ,
dus (afgerond) $x = 3,529$ , $x = 9,325$ of $x = ‐ 6,383$
$\cos (\frac{1}{10} x + 1) = ‐ 0,4$
$\to$ $\frac{1}{10} x + 1 = 1,9823... + k \cdot 2 π$ of $\frac{1}{10} x + 1 = 2 π - 1,9823... + k \cdot 2 π$ ( $k$ geheel) $\to$ $x = 9,823... + k \cdot 20 π$ of $x = 33,0089... + k \cdot 20 π$ ,
dus (afgerond) $x = 9,823$ , $x = 72,655$ of $x = 33,009$

c

Eerst $\sin (2 x) = 0,3$ $\to$ $2 x = 0,304... + k \cdot 2 π$ of $2 x = π - 0,304... + k \cdot 2 π$ ( $k$ geheel)
$\to$ $x = 0,1523... + k \cdot π$ of $x = 1,4184... + k \cdot π$ ,
dus oplossingen (afgerond) $x = 0,152$ , $x = 3,294$ , $x = 1,418$ en $x = 4,560$ ;
Schets maken met hulp GR en antwoord geven:
$0,152 < x < 1,418$ of $3,294 < x < 4,560$ .
Eerst $\cos (x + 1) = 0,15$ $\to$ $x + 1 = 1,420... + k \cdot 2 π$ of $x + 1 = 2 π - 1,420... + k \cdot 2 π$ (of $k$ geheel)
$\to$ $x = 0,420... + k \cdot 2 π$ of $x = 3,863... + k \cdot 2 π$ ,
dus oplossingen (afgerond) $x = 0,420$ , $x = 6,703$ en $x = 3,863$ ;
Schets maken met hulp GR en antwoord geven:
$0 \leq x < 0,420$ of $3,863 < x < 6,703$ .

17

a

$y = \frac{4 x + 3}{2 x - 1} = \frac{2 (2 x - 1) + 5}{2 x - 1} = \frac{2 (2 x - 1)}{2 x - 1} + \frac{5}{2 x - 1} = 2 + \frac{5}{2 x - 1}$ ;
$y = 2 + \frac{5}{2 x - 1} = 2 + 5 \cdot {(2 x - 1)}^{‐ 1}$ , dus $y' = - 5 \cdot {(2 x - 1)}^{‐ 2} \cdot 2 = \frac{‐ 10}{{(2 x - 1)}^{2}}$ .

b

$y = \frac{2}{x - 1} - \frac{2}{x + 1} = \frac{2 (x + 1)}{(x - 1) (x + 1)} - \frac{2 (x - 1)}{(x - 1) (x + 1)} = \frac{2 x + 2 - 2 x + 2}{x^{2} - 1} = \frac{4}{x^{2} - 1}$

c

$f (x) = \frac{x^{2} + 6}{2 x} = \frac{x^{2}}{2 x} + \frac{6}{2 x} = \frac{1}{2} x + 3 \cdot x^{‐ 1}$ , dus $f' (x) = \frac{1}{2} - 3 \cdot x^{‐ 2} = \frac{1}{2} - \frac{3}{x^{2}} = 0$ $\to$ $\frac{1}{2} = \frac{3}{x^{2}}$ $\to$ $x^{2} = 6$ , dus $x = ‐ \sqrt{6}$ of $x = \sqrt{6}$ .
$f (x) = \frac{x - 3}{2 x^{2}} = \frac{x}{2 x^{2}} - \frac{3}{2 x^{2}} = \frac{1}{2 x} - \frac{3}{2 x^{2}} = \frac{1}{2} \cdot \frac{1}{x} - \frac{3}{2} \cdot \frac{1}{x^{2}}$ , dus $f (x) = \frac{1}{2} \cdot x^{‐ 1} - \frac{3}{2} \cdot x^{‐ 2}$ ;
$f' (x) = ‐ \frac{1}{2} \cdot x^{‐ 2} + 3 \cdot x^{‐ 3} = \frac{‐ 1}{2 x^{2}} + \frac{3}{x^{3}} = \frac{‐ x}{2 x^{3}} + \frac{6}{2 x^{3}} = \frac{6 - x}{2 x^{3}} = 0$ $\to$ $x = 6$ .