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$ \sqrt 2 \cdot \sin (2x - \pi ) = 1 $ $ \sin (2x - \pi ) = \frac{1}{2}\sqrt 2 $ $ 2x - \pi = \frac{1}{4}\pi + k \cdot 2\pi $ of $ 2x - \pi = \frac{3}{4}\pi + k \cdot 2\pi $ $ 2x = 1\frac{1}{4}\pi + k \cdot 2\pi $ of $ 2x = 1\frac{3}{4}\pi + k \cdot 2\pi $ $ x = \frac{5}{8}\pi + k \cdot \pi $ of $ x = \frac{7}{8}\pi + k \cdot \pi $
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