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$\frac{3}{4}\text{π}$ en $1\frac{1}{4}\text{π}$

$1\frac{1}{3}\text{π}$ en $1\frac{2}{3}\text{π}$

$2\frac{1}{6}\text{π}$ en $2\frac{5}{6}\text{π}$

$t=\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $t=\frac{3}{4}\text{π}+k\cdot 2\text{π}$

$2\sin (x)=\sqrt{3}$ $\to$ $\sin (x)=\frac{1}{2}\sqrt{3}$ $\to$ $x=\frac{1}{3}\text{π}$ of $x=\text{π}-\frac{1}{3}\text{π}=\frac{2}{3}\text{π}$
periode $=2\text{π}$: $x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $x=\text{π}-\frac{1}{3}\text{π}+k\cdot 2\text{π}=\frac{2}{3}\text{π}+k\cdot 2\text{π}$;
Antwoord: $x=\frac{1}{3}\text{π}$ of $x=\frac{2}{3}\text{π}$

$\sin (2x)=\mathrm{‐}\frac{1}{2}\sqrt{3}$ $\to$ $2x=\mathrm{‐}\frac{1}{3}\text{π}$ of $2x=\text{π}-\mathrm{‐}\frac{1}{3}\text{π}=\frac{4}{3}\text{π}$ $\to$ $x=\mathrm{‐}\frac{1}{6}\text{π}$ of $x=\frac{2}{3}\text{π}$;
periode $=\text{π}$: $x=\mathrm{‐}\frac{1}{6}\text{π}+k\cdot \text{π}$ of $x=\frac{2}{3}\text{π}+k\cdot \text{π}$;
Antwoord: $x=\frac{5}{6}\text{π}$ of $x=1\frac{5}{6}\text{π}$ of $x=\frac{2}{3}\text{π}$ of $x=1\frac{2}{3}\text{π}$

$4\cos (2x)=2\sqrt{2}$ $\to$ $\cos (2x)=\frac{1}{2}\sqrt{2}$
$\to$ $2x=\frac{1}{4}\text{π}$ of $2x=2\text{π}-\frac{1}{4}\text{π}=\frac{7}{4}\text{π}$ $\to$ $x=\frac{1}{8}\text{π}$ of $x=\frac{7}{8}\text{π}$;
periode $=\text{π}$: $x=\frac{1}{8}\text{π}+k\cdot \text{π}$ of $x=\frac{7}{8}\text{π}+k\cdot \text{π}$;
Antwoord: $x=\frac{1}{8}\text{π}$ of $x=1\frac{1}{8}\text{π}$ of $x=\frac{7}{8}\text{π}$ of $x=1\frac{7}{8}\text{π}$

$1-2\sin (3x+\frac{1}{3}\text{π})=\mathrm{‐}1$ $\to$ $\sin (3x+\frac{1}{3}\text{π})=1$
$\to$ $3x+\frac{1}{3}\text{π}=\frac{1}{2}\text{π}$ $\to$ $x=\frac{1}{18}\text{π}$;
periode $=\frac{2}{3}\text{π}$: $x=\frac{1}{18}\text{π}+k\cdot \frac{2}{3}\text{π}$;
Antwoord: $x=\frac{1}{18}\text{π}$ of $x=\frac{13}{18}\text{π}$ of $x=\frac{25}{18}\text{π}$

$2\sqrt{2}-3\sin (\text{π}x)=\frac{1}{2}\sqrt{2}$ $\to$ $\sin (\text{π}x)=\frac{1}{2}\sqrt{2}$
$\to$ $\text{π}x=\frac{1}{4}\text{π}$ of $\text{π}x=\text{π}-\frac{1}{4}\text{π}=\frac{3}{4}\text{π}$ $\to$ $x=\frac{1}{4}$ of $x=\frac{3}{4}$;
periode $=2$: $x=\frac{1}{4}+k\cdot 2$ of $x=\frac{3}{4}+k\cdot 2$;
Antwoord: $x=\frac{1}{4}$ of $x=2\frac{1}{4}$ of $x=4\frac{1}{4}$ of $x=6\frac{1}{4}$ of $x=\frac{3}{4}$ of $x=2\frac{3}{4}$ of $x=4\frac{3}{4}$

$\sin (3x-\frac{1}{3}\text{π})=\frac{1}{2}\sqrt{3}$ $\to$ $3x-\frac{1}{3}\text{π}=\frac{1}{3}\text{π}$ of $3x-\frac{1}{3}\text{π}=\text{π}-\frac{1}{3}\text{π}=\frac{2}{3}\text{π}$
$\to$ $x=\frac{2}{9}\text{π}$ of $x=\frac{1}{3}\text{π}$;
periode $=\frac{2}{3}\text{π}$: $x=\frac{2}{9}\text{π}+k\cdot \frac{2}{3}\text{π}$ of $x=\frac{1}{3}\text{π}+k\cdot \frac{2}{3}\text{π}$;
Antwoord: $x=\frac{2}{9}\text{π}$ of $x=\frac{8}{9}\text{π}$ of $x=1\frac{5}{9}\text{π}$ of $x=\frac{1}{3}\text{π}$ of $x=\text{π}$ of $x=1\frac{2}{3}\text{π}$

${(\cos (x))}^2=\frac{1}{4}$ $\to$ $\cos (x)=‐\frac{1}{2}$ of $\cos (x)=\frac{1}{2}$ $\to$ ...
Alle oplossingen: $x=\frac{1}{3}\text{π}+k\cdot \text{π}$ of $x=\frac{2}{3}\text{π}+k\cdot \text{π}$;
Antwoord: $x=\frac{1}{3}\text{π}$ of $x=\frac{2}{3}\text{π}$ of $x=1\frac{1}{3}\text{π}$ of $x=1\frac{2}{3}\text{π}$