Period $2\pi$, amplitude $12$. Window: $[0,4\pi ]\times [-15,15]$.

Period $1$, amplitude $50$. Window: $[0,2]\times [-40,60]$.

Period $10$, amplitude $120$. Window $[0,20]\times [-120,120]$.

Period $\pi$, amplitude $20$. Window: $[0,2\pi ]\times [-20,20]$.

$cos(\frac{1}{2}x+4)=\frac{1}{5}$ gives $x=2\arccos \left(\frac{1}{5}\right)-8+k\cdot 4\pi \vee x=-2\arccos \left(\frac{1}{5}\right)-8+k\cdot 4\pi$ or $x\approx \mathrm{-5,261}+k\cdot 4\pi \vee x\approx \mathrm{-10,73}+k\cdot 4\pi$ .

$sin\left(\frac{\pi }{5}(x-2)\right)=\frac{1}{2}$ gives $x=\frac{5}{6}+2+k\cdot 10\vee x=\frac{25}{6}+2+k\cdot 10$.

$cos\left(4x\right)=\frac{1}{2}\sqrt{3}$ gives $x=\frac{1}{24}\pi +k\cdot \frac{1}{2}\pi \vee x=\mathrm{-}\frac{1}{24}\pi +k\cdot \frac{1}{2}\pi$.

$sin\left(\frac{2\pi }{15}x\right)=\frac{1}{6}$ gives $\frac{2\pi }{15}x=arcsin\left(\frac{1}{6}\right)+k\cdot 2\pi \vee \frac{\left(2\pi \right)}{\left(15\right)}x=\pi -arcsin\left(\frac{1}{6}\right)+k\cdot 2\pi$ or $x\approx \mathrm{0,399}+k\cdot 15\vee x\approx \mathrm{7,100}+k\cdot 15$.

${\text{B}}_f=[-10,30]$

$f\left(x\right)=0$ gives $x=\frac{8}{3}+k\cdot 8\vee x=-\frac{8}{3}+k\cdot 8$.
The zeroes are $(2\frac{2}{3},0),(5\frac{1}{3},0),(10\frac{2}{3},0)$ en $(13\frac{1}{3},0)$.

$2\frac{2}{3}\le x\le 5\frac{1}{3}\vee 10\frac{2}{3}\le x\le 13\frac{1}{3}$.

Enter: Y1=11+10*sin((2*pi/20)X) with window: $0\le X\le 20$ and $0\le Y\le 22$.

$11$ is the elevation of the axis of the wheel and $10$ is the radius of the wheel.

$40$ seconds

$h\left(t\right)=18$ gives $sin\left(\frac{2\pi }{20}t\right)=0,7$ and from that it follows that $t\approx \mathrm{2,468}+k\cdot 40\vee t\approx \mathrm{7,532}+k\cdot 40$.
So $\mathrm{5,1}$ seconds higher than $18$ m.