- Все ответы

$\sqrt{3}\sin x + \cos x < 1 \ \ \ | : 2$

$\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2} \cos x < \dfrac{1}{2}$

$\cos \dfrac{\pi}{6} \sin x + \sin \dfrac{\pi}{6} \cos x < \dfrac{1}{2}$

$\sin \left(\dfrac{\pi}{6} + x \right) < \dfrac{1}{2}$

$-\pi - \arcsin \dfrac{1}{2} + 2\pi n < \dfrac{\pi}{6} + x < \arcsin \dfrac{1}{2} + 2\pi n, \ n \in Z$

$-\pi - \dfrac{\pi}{6} + 2\pi n < \dfrac{\pi}{6} + x < \dfrac{\pi}{6} + 2\pi n, \ n \in Z$

$-\dfrac{7\pi}{6} + 2\pi n < \dfrac{\pi}{6} + x < \dfrac{\pi}{6} + 2\pi n, \ n \in Z \ \ \ \bigg| - \dfrac{\pi}{6}$

$-\dfrac{4\pi}{3} + 2\pi n < x < 2\pi n, \ n \in Z$

Ответ: $x \in \left(-\dfrac{4\pi}{3} + 2\pi n; \ 2\pi n \right), \ n \in Z$

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