Réponse :
si f(x) = 0 avec x ∈ R
alors demontrons que (1+cos(x) + sin(x)) ² - 2 (1+ cos(x))(1 +sin(x))= 0
autrement dit que (1+cos(x) + sin(x)) ² = 2 (1+ cos(x))
on a :
A = (1+cos(x) + sin(x)) ² = (1+cos(x) + sin(x))*(1+cos(x) + sin(x))
A = ((1+ cos(x) + sin(x))*(cos(x) + (1+ sin(x)))
A = (1+ cos(x)*cos(x) + (1+ cos(x)*(1+ sin(x)) +sin(x)*cos(x) + sin(x)*(1+ sin(x))
A = (1+cos(x)*(1+sin(x)) + (1+ cos(x)*cos(x) + sin(x)*cos(x) + sin(x)*(1+ sin(x))
A= (1+cos(x)*(1+sin(x)) + [cos(x) +cos(x)² + sin(x)*cos(x) +sin(x) +sin(x)²]
A = (1+cos(x)*(1+sin(x)) +[ cos(x) + 1 + sin(x)*(cos(x) +1)] car sin(x)²+cos(x)²=1
A = (1+cos(x)*(1+sin(x)) + [(cos(x) + 1)*(1 +sinx)]
A = (1+cos(x)*(1+sin(x)) + (1+cos(x)*(1+sin(x))
A = 2 (1+cos(x)*(1+sin(x))
donc l'égalité est vérifiée on a bien (1+cos(x) + sin(x)) ² = 2 (1+ cos(x))
et ainsi que (1+cos(x) + sin(x)) ² - 2 (1+ cos(x)) =0
j'espère avoir aidé