Please me! 1) 2cos(3x-п/3)=2 2) cos (п/4-x)=-1 3) - вопрос №12332241332603397624 от fedor1daniil12 08.07.2021 03:49

Question

Алгебра: Please me! 1) 2cos(3x-п/3)=2 2) cos (п/4-x)=-1 3) 3cos (2x-п/6)=0

fedor1daniil12 · Answer

1) 2cos(3x-П/3)=2cos(3x - π/3) =13x - π/3 = 2πk, k∈Z3x  =  π/3 + 2πk, k∈Zx  =  π/9 + 2πk/3, k∈Z2) cos (π/4 - x )= - 1cos (x - π/4 ) = - 1 x - π/4 = π + 2πn, n∈Zx = π + π/4 + 2πn, n∈Zx = 5π/4 + 2πn, n∈Z3) 3cos (2x - π/6) = 0cos (2x - π/6) = 02x = π/2  + π/6 + πm, m∈Z2x = 2π/3  + πm, m∈Zx = π/3  + πm/2, m∈Z...