Sin4x cosx-cos4x sinx больше √2/2

Решение: [latex]\sin4x\cos x - \cos4x\sin x \ \textgreater \ \frac{\sqrt2}{2} \\ \sin(4x-x) \ \textgreater \ \frac{\sqrt2}{2} \\ \sin3x \ \textgreater \ \frac{\sqrt2}{2} \\ \frac{3\pi}{4}+\pi k\ \textgreater \ 3x\ \textgreater \ \frac{\pi}{4} + \pi k \\ \\ \frac{\pi}{4} + \frac{\pi k}{3} \ \textgreater \ x \ \textgreater \ \frac{\pi}{12} + \frac{\pi k}{3}[/latex]