Bonjour,
cos(2x) + √3sin(2x) =-1
⇔ 1/2 x cos(2x) + √3/2 x sin(2x) = -1/2
⇔ sin(π/6)cos(2x) + cos(π/6)sin(2x) = -1/2
⇔ sin(π/6 + 2x) = -1/2 en utilisant sin(a+b) = sinacosb + sinbcosa
⇔ sin(π/6 + 2x) = sin(11π/6)
⇒ 2x + π/6 = 11π/6 + k2π
ou 2x + π/6 = π - 11π/6 + k2π
⇔ 2x = 10π/6 + k2π
ou 2x = -6π/6 + k2π
⇔ x = 5π/6 + kπ
ou x = -π/2 + kπ
