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$$\left.\sin^2x+\frac12\sin x-\frac{3}{50}=0\quad\right|+\frac{3}{50}$$$$\left.\sin^2x+\frac12\sin x=\frac{3}{50}\quad\right|+\frac{1}{16}$$$$\left.\sin^2x+\frac12\sin x+\frac{1}{16}=\frac{3}{50}+\frac{1}{16}\quad\right|\text{1-te binomische Formel}$$$$\left.\left(\sin x+\frac14\right)^2=\frac{24}{400}+\frac{25}{400}=\frac{49}{400}\quad\right|\sqrt{\cdots}$$$$\left.\sin x+\frac14=\pm\sqrt{\frac{49}{400}}=\pm\frac{7}{20}\quad\right|-\frac14$$$$\sin x=-\frac14\pm\frac{7}{20}=-\frac{5}{20}\pm\frac{7}{20}=\frac{-5\pm7}{20}$$$$\sin x=\left\{\begin{array}{l}-\frac{12}{20}\\[1ex]+\frac{2}{20}\end{array}\right\}=\left\{\begin{array}{l}-\frac{3}{5}\\[1ex]+\frac{1}{10}\end{array}\right.$$$$x=\left\{\begin{array}{l}2\pi\,n-\arcsin\left(\frac35\right)\\[1ex]2\pi\,n+\pi+\arcsin\left(\frac35\right)\\[1ex]2\pi\,n+\arcsin\left(\frac{1}{10}\right)\\[1ex]2\pi\,n+\pi-\arcsin\left(\frac{1}{10}\right)\end{array}\right.\quad n\in\mathbb Z$$
