Монгол Бодлогын Сан

Заавар: $2\sin\alpha\cos\alpha=\sin2\alpha$ томьёо ашиглан хялбарчил.

Бодолт: $$\sin\Big(x-\frac{\pi}{6}\Big)\cos\Big(x-\frac{\pi}{6}\Big)<\frac{\sqrt2}{4}\Leftrightarrow\sin\Big(x-\frac{\pi}{6}\Big)\cos\Big(x-\frac{\pi}{6}\Big)<\frac{\sqrt2}{2}\Leftrightarrow$$ $$\sin\Big(2x-\frac{\pi}{3}\Big)<\dfrac{\sqrt2}{2}$$ тул $$\dfrac{3\pi}{4}-2\pi+2\pi k<2x-\dfrac{\pi}{3}<\dfrac{\pi}{4}+2\pi k\Leftrightarrow-\dfrac{11\pi}{12}+2\pi k<2x<\dfrac{7\pi}{12}+2\pi k\Leftrightarrow$$ $$-\dfrac{11\pi}{24}+\pi k< x<\dfrac{7\pi}{24}+\pi k$$ тул $x\in\Big]k\pi-\dfrac{11\pi}{24};\dfrac{7\pi}{24}+k\pi\Big[$ байна.