Question

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

Answer 1

1) 2cos(3x-П/3)=2
cos(3x - π/3) =1
3x - π/3 = 2πk, k∈Z
3x  =  π/3 + 2πk, k∈Z
x  =  π/9 + 2πk/3, k∈Z

2) cos (π/4 - x )= - 1
cos (x - π/4 ) = - 1
 x - π/4 = π + 2πn, n∈Z
x = π + π/4 + 2πn, n∈Z
x = 5π/4 + 2πn, n∈Z

3) 3cos (2x - π/6) = 0
cos (2x - π/6) = 0
2x = π/2  + π/6 + πm, m∈Z
2x = 2π/3  + πm, m∈Z
x = π/3  + πm/2, m∈Z

Answer 2

Спасибо большое!