a) cos(3x)=1/2 => 3x = π/3 +2kπ ou 3x = -π/3+2k'π => x=π/9 +(2/3)kπ ou x = -π/9+(2/3)k'π
On restreint les solutions à [0;2π] en ajustant les valeurs de k et k', et on trouve que les solutions sont :
{π/9; 7π/9 ; 13π/9 ; 11π/9 ; 5π/9 ; 17π/9}
b) sin(2x-π/4)=0 => 2x- π/4 = 0 + kπ => 2x = π/4 + kπ => x = π/8 + kπ/2
on restreint les solutions à ]-π; π] et on trouve :
{ π/8; 5π/8 ; -3π/8 ; -7π/8}
c) 2cos(π - x/2) = -1 => cos(π-x/2) = -1/2
=> π-x/2 = 2π/3 + 2kπ ou π-x/2 = -2π/3 +2k'π
=> x/2 = π-2π/3 - 2kπ ou x/2 = π+2π/3 - 2k'π
=> x = 2π/3 -4kπ ou x = 10π/3 - 4kπ
les solutions sur ]-2π;2π] sont donc :
{-2π/3 ; 2π/3}
