Réponse:
cos(2x) = ½ <=> cos(2x) = cos(π/3)
2x = π/3 + 2kπ ou 2x = -π/3 + 2kπ avec k entier relatif
x = π/6 + kπ ou x = -π/6 + kπ
Dans IR, S ={-π/6 + kπ ; π/6+kπ , k€ Z }
Dans ]-π;π] ( autrement dit dans ]-6π/6; 6π/6] )
S = { -5π/6; -π/6; π/6; 5π/6}
1a On obtient :
4X² - 2(1+√3)X + √3 = 0 (2)
1b
∆ = [- 2(1+√3)]² - 4×4×√3
∆ = 4(1+√3)² - 16√3
∆ = 4(1+2√3+3)-16√3
∆ = 4+8√3+12-16√3
∆ = 4-8√3+12
∆ = 4(1-2√3+3)
∆ = 4(1-√3)²
1c
∆ > 0
√∆ = √[4(1-√3)²] = 2√(1-√3)² = 2×|1-√3| = 2(√3-1) car 1-√3 <0
L'equation (2) admet 2 solutions :
X1 =[ 2(1+√3)-2(√3-1)] / 8
X1 = 4/8
X1 = 1/2
X2 = [ 2(1+√3)+ 2(√3-1)] / 8
X2 = 4√3/8
X2 = √3/2
S = { 1/2 ; √3/2}
2.
Dans ]-π;π] :
cos(x) = ½ <=>
x = π/3 ou x= -π/3
cos(x) = √3/2 <=>
x = π/6 ou x = -π/6
S ={-π/3; -π/6; π/6; π/3}
Dans IR :
S ={-π/3+2kπ; -π/6+2kπ; π/6+2kπ; π/3+2kπ, k€ Z }
