Question

Помогите с тригонометрией

Answer 1

$\sin\left(\dfrac{\pi}{3} - 2x \right) = -2\cos^{2}\left(\dfrac{\pi}{12} + x \right) - 1$

$\sin\dfrac{\pi}{3} \cos 2x - \cos\dfrac{\pi}{3}\sin 2x = -\left(2\cos^{2}\left(\dfrac{\pi}{12} + x \right) - 1 + 2\right)$

$\dfrac{\sqrt{3}}{2}\cos 2x - \dfrac{1}{2}\sin 2x = -\left(\cos \left(\dfrac{\pi}{6} + 2x \right) + 2 \right)$

$\dfrac{\sqrt{3}}{2}\cos 2x - \dfrac{1}{2}\sin 2x = -\cos\left(\dfrac{\pi}{6} + 2x \right) - 2$

$\dfrac{\sqrt{3}}{2}\cos 2x - \dfrac{1}{2}\sin 2x = \sin \dfrac{\pi}{6}\sin 2x - \cos \dfrac{\pi}{6}\cos 2x - 2$

$\dfrac{\sqrt{3}}{2}\cos 2x - \dfrac{1}{2}\sin 2x = \dfrac{1}{2} \sin 2x - \dfrac{\sqrt{3}}{2}\cos 2x - 2$

$\sqrt{3}\cos 2x - \sin 2x = -2 \ \ \ | : \dfrac{1}{\sqrt{(\sqrt{3})^{2} + (-1)^{2}}}$

$\dfrac{\sqrt{3}}{2}\cos 2x - \dfrac{1}{2} \sin 2x = -1$

$\cos \dfrac{\pi}{6} \cos 2x - \sin\dfrac{\pi}{6}\sin 2x = -1$

$\cos \left(\dfrac{\pi}{6} + 2x \right) = -1$

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

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

$2x = \dfrac{5\pi}{6} + 2\pi n, \ n \in Z$

$x = \dfrac{5\pi}{12} + \pi n, \ n \in Z$

Ответ: $x = \dfrac{5\pi}{12} + \pi n, \ n \in Z$