Bonjour,
1) tan(2x + π/4) = 1
⇔ 2x + π/4 = π/4 + kπ
⇔ x = kπ/2
Sur ]-π;2π] :
k = -2 ⇒ x = -π exclus
k = -1 ⇒ x = -π/2
k = 0 ⇒ x = 0
k = 1 ⇒ x = π/2
k = 2 ⇒ x = π
k = 3 ⇒ x = 3π/2
k = 4 ⇒ x = 2π
2) sin(2x + π/3) = cos(3x)
⇔ sin(2x + π/3) = sin(3x + π/2)
⇔ 2x + π/3 = 3x + π/2 + k2π
ou 2x + π/3 = π - (3x + π/2) + k2π
⇔ x = π/3 - π/2 - k2π
ou x = (-π/3 + π/2 + k2π)/5
⇔ x = -π/6 - k2π
ou x = π/30 + k2π/5
Sur ]-π;2π] : idem on remplace k pour trouver les solutions appartenant à cet intervalle
3) cos(2x - 3π/4) = √3/2
⇔ cos(2x - 3π/4) = cos(π/6)
⇔ 2x - 3π/4 = π/6 + k2π
ou 2x - 3π/4 = -π/6 + k2π
⇔ x = π/12 + 3π/8 + kπ
ou x = -π/12 + 3π/8 + kπ
...
4) sin²(x) - sin(x)cos(x) = 0
⇔ sin(x)[sin(x) - cos(x)] = 0
⇔ sin(x) = 0
ou sin(x) = cos(x)
⇔ x = kπ
ou x = π/4 + kπ
