Ответы на вопрос:
2cosx - 1 = 0 [0; 2pi] 2cosx = 1 cosx = 1/2 x = +- pi/3 + 2pik, k ∈ z 1) k = -1 x1 = + pi/3 - 2pi = pi/3 - 6pi/3 = (pi-6pi)/3 = -5pi/3 ∉ x2 = - pi/3 - 2pi = - pi/3 - 6pi/3 = (-pi-6pi)/3 = -7pi/3 ∉ 2) k = 0 x1 = +pi/3 ∈ x2 = - pi/3 ∉ 3) k =1 x1 = +pi/3 + 2pi = pi/3 + 6pi/3 = (pi+6pi)/3 = 7pi/3 ∉ x2 = -pi/3 + 2pi = -pi/3 + 6pi/3 = (-pi+6pi)/3 = 5pi/3 ∈ 4) k = 2 x1 = +pi/3 + 4pi = pi/3 + 12pi/3 = (pi+12pi)/3 = 13pi/3 ∉ x2 = -pi/3 + 4pi = -pi/3 + 12pi/3 = (-pi+12pi)/3 = 11pi/3 ∉ ответ: pi/3; 5pi/3
Пошаговое объяснение:
выражение
(3х+7)-4х
при х=-6
(3*(-6+7)-4х(-6)=(-18+7)-(-24)=-11+24=13
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