Делим сначала неравенство на положительное число 2, получим [tex]\frac{1}{2}cosx-\frac{\sqrt3}{2}sinx<\frac{\sqrt2}{2}\\sin\frac{\pi}{6}cosx-cos\frac{\pi}{6}sinx<\frac{\sqrt2}{2}\\sin(\frac{\pi}{6}-x)<\frac\sqrt2}{2}\\\frac{\pi}{4}+2\pi n<\frac{\pi}{6}-x<\frac{3\pi}{4}+2\pi n , \; n\in Z\\-\frac{7\pi}{12}+2\pi n
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[tex]\frac{1}{2}cosx-\frac{\sqrt3}{2}sinx<\frac{\sqrt2}{2}\\sin\frac{\pi}{6}cosx-cos\frac{\pi}{6}sinx<\frac{\sqrt2}{2}\\sin(\frac{\pi}{6}-x)<\frac\sqrt2}{2}\\\frac{\pi}{4}+2\pi n<\frac{\pi}{6}-x<\frac{3\pi}{4}+2\pi n , \; n\in Z\\-\frac{7\pi}{12}+2\pi n
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Делим сначала неравенство на положительное число 2, получим[tex]\frac{1}{2}cosx-\frac{\sqrt3}{2}sinx<\frac{\sqrt2}{2}\\sin\frac{\pi}{6}cosx-cos\frac{\pi}{6}sinx<\frac{\sqrt2}{2}\\sin(\frac{\pi}{6}-x)<\frac\sqrt2}{2}\\\frac{\pi}{4}+2\pi n<\frac{\pi}{6}-x<\frac{3\pi}{4}+2\pi n , \; n\in Z\\-\frac{7\pi}{12}+2\pi n