Монгол Бодлогын Сан

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

Бодолт: $$\sin\Big(3x-\frac{\pi}{6}\Big)\cos\Big(3x-\frac{\pi}{6}\Big)<\frac{\sqrt2}{4}\Leftrightarrow2\sin\Big(3x-\frac{\pi}{6}\Big)\cos\Big(3x-\frac{\pi}{6}\Big)<\frac{\sqrt2}{2}\Leftrightarrow$$ $$\sin\Big(6x-\frac{\pi}{3}\Big)<\dfrac{\sqrt2}{2}$$ тул $$\dfrac{3\pi}{4}+2\pi (k-1)<6x-\dfrac{\pi}{3}<\dfrac{\pi}{4}+2\pi k\Leftrightarrow-\dfrac{11\pi}{12}+2\pi k<6x<\dfrac{7\pi}{12}+2\pi k\Leftrightarrow$$ $$-\dfrac{11\pi}{72}+\dfrac{\pi k}{3}< x<\dfrac{7\pi}{72}+\dfrac{\pi k}{3}$$ тул $x\in\left]\dfrac{k\pi}{3}-\dfrac{11\pi}{72};\dfrac{7\pi}{72}+\dfrac{k\pi}{3}\right[$ байна.