Question

найти нули функции y=-2cos(3x-п/6)

Answer 1

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