$\sin (x)=\sin (\frac{1}{3}\text{π})$
$x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $x=\text{π}-\frac{1}{3}\text{π}+k\cdot 2\text{π}=\frac{2}{3}\text{π}+k\cdot 2\text{π}$
De oplossingen zijn: $\frac{1}{3}\text{π}$ en $\frac{2}{3}\text{π}$.

$\cos (x)=\cos (\frac{1}{3}\text{π})$
$x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{1}{3}\text{π}+k\cdot 2\text{π}$
De oplossingen zijn: $\frac{1}{3}\text{π}$ en $1\frac{2}{3}\text{π}$.

$\sin (x)=\sin (4x)$
$x=4x+k\cdot 2\text{π}$ of $x=\text{π}-4x+k\cdot 2\text{π}$
$\mathrm{‐}3x=k\cdot 2\text{π}$ of $5x=\text{π}+k\cdot 2\text{π}$
$x=k\cdot \mathrm{‐}\frac{2}{3}\text{π}$ of $x=\frac{1}{5}\text{π}+k\cdot \frac{2}{5}\text{π}$
De oplossingen zijn: $0$, $\frac{2}{3}\text{π}$, $1\frac{1}{3}\text{π}$, $\frac{1}{5}\text{π}$, $\frac{3}{5}\text{π}$, $\text{π}$, $1\frac{2}{5}\text{π}$, $1\frac{4}{5}\text{π}$ en $2\text{π}$.

$\cos (x)=\cos (4x)$
$x=4x+k\cdot 2\text{π}$ of $x=\mathrm{‐}4x+k\cdot 2\text{π}$
$\mathrm{‐}3x=k\cdot 2\text{π}$ of $5x=k\cdot 2\text{π}$
$x=k\cdot \mathrm{‐}\frac{2}{3}\text{π}$ of $x=k\cdot \frac{2}{5}\text{π}$
De oplossingen zijn: $0$, $\frac{2}{3}\text{π}$, $1\frac{1}{3}\text{π}$, $2\text{π}$, $\frac{2}{5}\text{π}$, $\frac{4}{5}\text{π}$, $1\frac{1}{5}\text{π}$ en $1\frac{3}{5}\text{π}$.

$\sin (x)=\mathrm{‐}\sin (x)$
$\sin (x)=\sin (\mathrm{‐}x)$
$x=\mathrm{‐}x+k\cdot 2\text{π}$ of $x=\text{π}-\mathrm{‐}x+k\cdot 2\text{π}$
$2x=k\cdot 2\text{π}$ dus $x=k\cdot \text{π}$
De oplossingen zijn: $0$, $\text{π}$ en $2\text{π}$.

$\cos (x)=\mathrm{‐}\cos (x)$
$\cos (x)=\cos (x+\text{π})$
$x=x+\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}x-\text{π}+k\cdot 2\text{π}$
$2x=\mathrm{‐}\text{π}+k\cdot 2\text{π}$ dus $x=\mathrm{‐}\frac{1}{2}\text{π}+k\cdot \text{π}$
De oplossingen zijn: $\frac{1}{2}\text{π}$ en $1\frac{1}{2}\text{π}$.

$\sin (x)=\cos (\frac{1}{3}\text{π})$
$\cos (\frac{1}{2}\text{π}-x)=\cos (\frac{1}{3}\text{π})$
$\frac{1}{2}\text{π}-x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-x=\mathrm{‐}\frac{1}{3}\text{π}+k\cdot 2\text{π}$
$x=\frac{1}{6}\text{π}-k\cdot 2\text{π}$ of $x=\frac{5}{6}\text{π}-k\cdot 2\text{π}$
De oplossingen zijn: $\frac{1}{6}\text{π}$ en $\frac{5}{6}\text{π}$.

$\cos (x)=\sin (\frac{1}{3}\text{π})$
$\sin (\frac{1}{2}\text{π}-x)=\sin (\frac{1}{3}\text{π})$
$\frac{1}{2}\text{π}-x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-x=\text{π}-\frac{1}{3}\text{π}+k\cdot 2\text{π}$
$x=\frac{1}{6}\text{π}-k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{1}{6}\text{π}-k\cdot 2\text{π}$
De oplossingen zijn: $\frac{1}{6}\text{π}$ en $1\frac{5}{6}\text{π}$.

$\sin (x)=\cos (x)$
$\cos (\frac{1}{2}\text{π}-x)=\cos (x)$
$\frac{1}{2}\text{π}-x=x+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-x=\mathrm{‐}x+k\cdot 2\text{π}$
$2x=\frac{1}{2}\text{π}-k\cdot 2\text{π}$ dus $x=\frac{1}{4}\text{π}-k\cdot \text{π}$
De oplossingen zijn: $\frac{1}{4}\text{π}$ en $1\frac{1}{4}\text{π}$.

$\sin (x)=\mathrm{‐}\cos (x)$
$\cos (\frac{1}{2}\text{π}-x)=\cos (\text{π}-x)$
$\frac{1}{2}\text{π}-x=\text{π}-x+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-x=\mathrm{‐}\text{π}+x+k\cdot 2\text{π}$
$2x=1\frac{1}{2}\text{π}-k\cdot 2\text{π}$ dus $x=\frac{3}{4}\text{π}-k\cdot \text{π}$
De oplossingen zijn: $\frac{3}{4}\text{π}$ en $1\frac{3}{4}\text{π}$.

$\sin (x)=\sin (x+2)$
$x=x+2+k\cdot 2\text{π}$ of $x=\text{π}-(x+2)+k\cdot 2\text{π}$
$2x=\text{π}-2+k\cdot 2\text{π}$ dus $x=\frac{1}{2}\text{π}-1+k\cdot \text{π}$
De oplossingen zijn: $\frac{1}{2}\text{π}-1$ en $1\frac{1}{2}\text{π}-1$.

$\cos (x)=\cos (x+2)$
$x=x+2+k\cdot 2\text{π}$ of $x=\mathrm{‐}(x+2)+k\cdot 2\text{π}$
$2x=\mathrm{‐}2+k\cdot 2\text{π}$ dus $x=\mathrm{‐}1+k\cdot \text{π}$
De oplossingen zijn: $\mathrm{‐}1+\text{π}$ en $\mathrm{‐}1+2\text{π}$.

$x=3x-2\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $x=\text{π}-(3x-2\frac{1}{4}\text{π)}+k\cdot 2\text{π}$
$\mathrm{‐}2x=\mathrm{‐}2\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $4x=3\frac{1}{4}\text{π}+k\cdot 2\text{π}$
$x=1\frac{1}{8}\text{π}-k\cdot \text{π}$ of $x=\frac{13}{16}\text{π}+k\cdot \frac{1}{2}\text{π}$
De oplossingen zijn: $\frac{1}{8}\text{π}$, $1\frac{1}{8}\text{π}$, $\frac{5}{16}\text{π}$, $\frac{13}{16}\text{π}$, $1\frac{5}{16}\text{π}$ en $1\frac{13}{16}\text{π}$.

$x+\frac{1}{4}\text{π}=2x-1\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $x+\frac{1}{4}\text{π}=\mathrm{‐}(2x-1\frac{1}{4}\text{π)}+k\cdot 2\text{π}$
$\mathrm{‐}x=\mathrm{‐}1\frac{1}{2}\text{π}+k\cdot 2\text{π}$ of $3x=\text{π}+k\cdot 2\text{π}$
$x=1\frac{1}{2}\text{π}-k\cdot 2\text{π}$ of $x=\frac{1}{3}\text{π}+k\cdot \frac{2}{3}\text{π}$
De oplossingen zijn: $1\frac{1}{2}\text{π}$, $\frac{1}{3}\text{π}$, $\text{π}$ en $1\frac{2}{3}\text{π}$.

$\sin (x-\frac{5}{6}\text{π})=\sin (\frac{1}{2}\text{π}-x)$
$x-\frac{5}{6}\text{π}=\frac{1}{2}\text{π}-x+k\cdot 2\text{π}$ of $x-\frac{5}{6}\text{π}=\text{π}-(\frac{1}{2}\text{π}-x)+k\cdot 2\text{π}$
$2x=1\frac{1}{3}\text{π}+k\cdot 2\text{π}$ dus $x=\frac{2}{3}\text{π}+k\cdot \text{π}$
De oplossingen zijn: $\frac{2}{3}\text{π}$ en $1\frac{2}{3}\text{π}$.

$\cos (\frac{1}{2}\text{π}-x)=\cos (\frac{6}{7}\text{π})$
$\frac{1}{2}\text{π}-x=\frac{6}{7}\text{π}+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-x=\mathrm{‐}\frac{6}{7}\text{π}+k\cdot 2\text{π}$
$x=\mathrm{‐}\frac{5}{14}\text{π}-k\cdot 2\text{π}$ of $x=1\frac{5}{14}\text{π}-k\cdot 2\text{π}$
De oplossingen zijn: $1\frac{9}{14}\text{π}$ en $1\frac{5}{14}\text{π}$.

$\sin (x+\frac{1}{4}\text{π})=\cos (\text{π}+x)$
$\sin (x+\frac{1}{4}\text{π})=\sin (\mathrm{‐}\frac{1}{2}\text{π}-x)$
$x+\frac{1}{4}\text{π}=\mathrm{‐}\frac{1}{2}\text{π}-x+k\cdot 2\text{π}$ of $x+\frac{1}{4}\text{π}=\text{π}-(\mathrm{‐}\frac{1}{2}\text{π}-x)+k\cdot 2\text{π}$
$2x=\mathrm{‐}\frac{3}{4}\text{π}+k\cdot 2\text{π}$ dus $x=\mathrm{‐}\frac{3}{8}\text{π}+k\cdot \text{π}$
De oplossingen zijn: $\frac{5}{8}\text{π}$ en $1\frac{5}{8}\text{π}$.

$\cos (\frac{1}{2}\text{π}-2x)=\cos (3x)$
$\frac{1}{2}\text{π}-2x=3x+k\cdot 2\text{π}$ of $\frac{1}{2}\text{π}-2x=\mathrm{‐}3x+k\cdot 2\text{π}$
$\mathrm{‐}5x=\mathrm{‐}\frac{1}{2}\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{1}{2}\text{π}+k\cdot 2\text{π}$
$x=\frac{1}{10}\text{π}-k\cdot \frac{2}{5}\text{π}$ of $x=\mathrm{‐}\frac{1}{2}\text{π}+k\cdot 2\text{π}$
De oplossingen zijn: $\frac{1}{10}\text{π}$, $\frac{1}{2}\text{π}$, $\frac{9}{10}\text{π}$, $1\frac{3}{10}\text{π}$, $1\frac{7}{10}\text{π}$ en $1\frac{1}{2}\text{π}$.

$1-{\sin }^2(x)+\sin (x)=1$
${\sin }^2(x)-\sin (x)=0$
$\sin (x)(\sin (x)-1)=0$
$\sin (x)=0$ of $\sin (x)=1$
$x=k\cdot \text{π}$ of $x=\frac{1}{2}\text{π}+k\cdot 2\text{π}$

$2{\cos }^2(x)-1+\cos (x)+1=0$
$2{\cos }^2(x)+\cos (x)=0$
$2\cos (x)(\cos (x)+\frac{1}{2})=0$
$2\cos (x)=0$ of $\cos (x)=\mathrm{‐}\frac{1}{2}$
$x=\frac{1}{2}\text{π}+k\cdot \text{π}$ of $x=\frac{2}{3}\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{2}{3}\text{π}+k\cdot 2\text{π}$

$1-{\sin }^2(x)+3\sin (x)=3$
${\sin }^2(x)-3\sin (x)+2=0$
$(\sin (x)-2)(\sin (x)-1)=0$
$\sin (x)=2$ of $\sin (x)=1$
$x=\frac{1}{2}\text{π}+k\cdot 2\text{π}$

$1-2{\sin }^2(x)=1-\sin (x)$
$2{\sin }^2(x)-\sin (x)=0$
$2\sin (x)(\sin (x)-\frac{1}{2})=0$
$\sin (x)=0$ of $\sin (x)=\frac{1}{2}$
$x=k\cdot \text{π}$ of $x=\frac{1}{6}\text{π}+k\cdot 2\text{π}$ of $x=\frac{5}{6}\text{π}+k\cdot 2\text{π}$

$\frac{\sin (x+\frac{1}{3}\text{π})}{\sin (x-\frac{1}{3}\text{π})}=\frac{\sin (x)\cdot \cos (\frac{1}{3}\text{π})+\cos (x)\cdot \sin (\frac{1}{3}\text{π})}{\sin (x)\cdot \cos (\frac{1}{3}\text{π})-\cos (x)\cdot \sin (\frac{1}{3}\text{π})}=$
$\frac{\frac{1}{2}\sin (x)+\frac{1}{2}\sqrt{3}\cos (x)}{\frac{1}{2}\sin (x)-\frac{1}{2}\sqrt{3}\cos (x)}=\frac{\sin (x)+\sqrt{3}\cos (x)}{\sin (x)-\sqrt{3}\cos (x)}$

$\frac{{\left(\sin (x)+\cos (x)\right)}^2}{{\left(\sin (x)-\cos (x)\right)}^2}=\frac{{\sin }^2(x)+{\cos }^2(x)+2\sin (x)\cdot \cos (x)}{{\sin }^2(x)+{\cos }^2(x)-2\sin (x)\cdot \cos (x)}=\frac{1+\sin (2x)}{1-\sin (2x)}$

$\frac{\cos (2x)}{\sin (x)+\cos (x)}=\frac{{\cos }^2(x)-{\sin }^2(x)}{\sin (x)+\cos (x)}=\frac{\left(\cos (x)-\sin (x)\right)\left(\cos (x)+\sin (x)\right)}{\sin (x)+\cos (x)}=\cos (x)-\sin (x)$

$\frac{\cos (2x)}{\cos (x)}=\frac{2{\cos }^2(x)-1}{\cos (x)}=2\cos (x)-\frac{1}{\cos (x)}$

$\frac{\cos (2(x-\frac{1}{4}\text{π}))}{\cos (x-\frac{1}{2}\text{π})}=\frac{\cos (2x-\frac{1}{2}\text{π})}{\cos (x-\frac{1}{2}\text{π})}=\frac{\sin (2x)}{\sin (x)}=\frac{2\cdot \sin (x)\cdot \cos (x)}{\sin (x)}=2\cdot \cos (x)$

${\sin }^2(x)+2\sin (x)\cos (x)+{\cos }^2(x)=1+\cos (x)$
$2\sin (x)\cos (x)=\cos (x)$
$\cos (x)=0$ of $2\sin (x)=1$
$\cos (x)=0$ of $\sin (x)=\frac{1}{2}$
$x=\frac{1}{2}\text{π}+k\cdot \text{π}$ of $x=\frac{1}{6}\text{π}+k\cdot 2\text{π}$ of $x=\frac{5}{6}\text{π}+k\cdot 2\text{π}$

$2{\cos }^2(x)+2\sin (x)\cos (x)=0$
$\cos (x)(\cos (x)+\sin (x))=0$
$\cos (x)=0$ of $\cos (x)=\mathrm{‐}\sin (x)$
$x=\frac{1}{2}\text{π}+k\cdot \text{π}$ of $x=\mathrm{‐}\frac{1}{4}\text{π}+k\cdot \text{π}$

$2{\sin }^2(x)-1=\mathrm{‐}\cos (x+\frac{1}{4}\text{π})$
$\mathrm{‐}\cos (2x)=\mathrm{‐}\cos (x+\frac{1}{4}\text{π})$
$2x=x+\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $2x=\mathrm{‐}(x+\frac{1}{4}\text{π})+k\cdot 2\text{π}$
$x=\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $3x=\mathrm{‐}\frac{1}{4}\text{π}+k\cdot 2\text{π}$
$x=\frac{1}{4}\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{1}{12}\text{π}+k\cdot \frac{2}{3}\text{π}$

$4{\sin }^3(x)=3\cdot 2\sin (x)\cos (x)$
$4{\sin }^3(x)-6\sin (x)\cos (x)=0$
$\sin (x)(4{\sin }^2(x)-6\cos (x))=0$
$\sin (x)=0$ of $4{\sin }^2(x)=6\cos (x)$
$\sin (x)=0$ of $4-4{\cos }^2(x)=6\cos (x)$
$\sin (x)=0$ of $4{\cos }^2(x)+6\cos (x)-4=0$
$\sin (x)=0$ of $(2\cos (x)+4)(2\cos (x)-1)=0$
$\sin (x)=0$ of $\cos (x)=\mathrm{‐}2$ of $\cos (x)=\frac{1}{2}$
$x=k\cdot \text{π}$ of $x=\frac{1}{3}\text{π}+k\cdot 2\text{π}$ of $x=\mathrm{‐}\frac{1}{3}\text{π}+k\cdot 2\text{π}$

$(\cos (x)\cdot \frac{1}{2}\sqrt{2}-\sin (x)\cdot \frac{1}{2}\sqrt{2})(\cos (x)\cdot \frac{1}{2}\sqrt{2}+\sin (x)\cdot \frac{1}{2}\sqrt{2})=\sin (x)\cos (x)$
${\cos }^2(x)-{\sin }^2(x)=2\sin (x)\cos (x)$
$\cos (2x)=\sin (2x)$
$x=\frac{1}{8}\text{π}+k\cdot \frac{1}{2}\text{π}$

${\cos }^4(x)-{\sin }^4(x)=({\cos }^2(x)-{\sin }^2(x))({\cos }^2(x)+{\sin }^2(x))=$
$({\cos }^2(x)-{\sin }^2(x))\cdot 1={\cos }^2(x)-{\sin }^2(x)$

${\sin }^6(x)+3{\sin }^2(x)\cdot {\cos }^2(x)+{\cos }^6(x)=$
${\sin }^6(x)+3{\sin }^2(1-{\sin }^2(x))+{(1-{\sin }^2(x))}^3=$
${\sin }^6(x)+3{\sin }^2(x)-3{\sin }^4(x)+1-3{\sin }^2(x)+3{\sin }^4(x)-{\sin }^6(x)=1$

$\sin (x+y)\cdot \sin (x-y)=$
$(\sin (x)\cos (y)+\cos (x)\sin (y))(\sin (x)\cos (y)-\cos (x)\sin (y))=$
${\sin }^2(x){\cos }^2(y)-{\cos }^2(x){\sin }^2(y)=$
$(1-{\cos }^2(x)){\cos }^2(y)-{\cos }^2(x)(1-{\cos }^2(y))=$
${\cos }^2(y)-{\cos }^2(x)\cdot {\cos }^2(y)-{\cos }^2(x)+{\cos }^2(x)\cdot {\cos }^2(y)={\cos }^2(y)-{\cos }^2(x)$,
dus ${\cos }^2(y)-\sin (x+y)\sin (x-y)-{\cos }^2(x)=0$