Question

Sin 2x = корень3sin(3п/2 - х)
Промежуток [3п; 4п]

Answer 1

$\sin(2x) = \sqrt{3} \sin( \frac{3\pi}{2} - x ) \\ \sin(2x) = - \sqrt{3} \cos(x) \\ 2 \sin(x) \cos(x) + \sqrt{3} \cos(x) = 0 \\ \cos(x) (2 \sin(x) + \sqrt{3} ) = 0 \\ \\ \cos(x) = 0 \\ x1 = \frac{\pi}{2} + \pi \: n \\ \\ \sin(x) = - \frac{ \sqrt{3} }{2} \\ x2 = - \frac{\pi}{3} + 2\pi \: n \\ x3 = - \frac{2\pi}{3} + 2\pi \: n$

на промежутке:

рисунок

$x1 = 3\pi + \frac{\pi}{3} = \frac{10\pi}{3} \\ x2 = 4\pi - \frac{\pi}{3} = \frac{11\pi}{3}$

$a)x1 = \frac{\pi}{2} + \pi \: n \\ x2 = - \frac{\pi}{3} + 2\pi \: n \\ x3 = - \frac{2\pi}{3} + 2 \pi \: n \\ b) \frac{10\pi}{3} ; \frac{7\pi}{2} ; \frac{11\pi}{3}$

n принадлежит Z.