$\sin (\mathrm{x})=\cos (\mathrm{x})-1\backslash sin(\backslash mathrm\; x)=\backslash cos(\backslash mathrm\; x)-1$

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Suche gleiche Funktionswerte.

$\begin{array}{c}\sin \left(0\right)=\cos \left(0\right)-1=0\\ \sin \left(\frac{3\pi }{2}\right)=\cos \left(\frac{3\pi }{2}\right)-1=-1\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\backslash sin\backslash left(0\backslash right)=\backslash cos\backslash left(0\backslash right)-1=0\backslash \backslash \backslash sin\backslash left(\backslash frac\{3\backslash pi\}2\backslash right)=\backslash cos\backslash left(\backslash frac\{3\backslash pi\}2\backslash right)-1=-1\backslash end\{array\}$

$x=0x=0$ und $x=\frac{3\pi }{2}x=\backslash frac\{3\backslash pi\}2$ sind Lösungen. Periodizität beachten.

Lösungen: $x_1=0+2\mathrm{k}\pi ,x_2=\frac{3\pi }{2}+2\mathrm{k}\pi ,k\in \mathbb{Z}x\_1=0\backslash ;+\backslash ;2\backslash mathrm\; k\backslash mathrm\backslash pi,\backslash ;x\_2=\backslash frac\{3\backslash mathrm\backslash pi\}2+2\backslash mathrm\; k\backslash mathrm\backslash pi,\backslash ;k\backslash in\backslash mathbb\{Z\}$

Anhand von Funktionsgraphen kann man erkennen, dass es keine weiteren Lösungen gibt.