Question

найти нули функции y=-2cos(3x-п/6)

Answer 1

$y = - 2( \frac{ \sqrt{3} }{2} \sin(3x) + \frac{1}{2} \cos(3x) ) = - \sqrt{3} \sin(3x) - \cos(3x) = 0 \\ - \sqrt{3} = \frac{ \cos(3x) }{ \sin(3x) } \\ - \sqrt{3} = \cot(3x) \\ 3x = \frac{2\pi}{3} + \pi \times k \\ x = \frac{2\pi}{9} + \frac{\pi \times k}{3}$