Quiz — 1616425705_18Equations cos(x + a) = α.pdf

Question 1 sur 5 10:00

[{"id":283,"question":"Quelle est la solution générale de l'équation cos(x) = 1\/2 dans l'intervalle [0, 2π] ?","option_a":"x = π\/3 + 2kπ ou x = 5π\/3 + 2kπ, k ∈ ℤ","option_b":"x = π\/6 + 2kπ ou x = 11π\/6 + 2kπ, k ∈ ℤ","option_c":"x = π\/3 + kπ, k ∈ ℤ","option_d":"x = π\/6 + kπ, k ∈ ℤ","option_e":"","option_f":"","bonne_reponse":"A","explication":"Les solutions de cos(x) = 1\/2 sont x = π\/3 + 2kπ et x = 5π\/3 + 2kπ (ou -π\/3 + 2kπ) car le cosinus est positif dans les premier et quatrième quadrants.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":284,"question":"Si cos(x + π\/4) = √2\/2, quelle est la valeur de x dans [0, 2π] ?","option_a":"x = π\/4 ou x = 7π\/4","option_b":"x = 0 ou x = π\/2","option_c":"x = π\/2 ou x = 3π\/2","option_d":"x = π\/4 ou x = 3π\/4","option_e":"","option_f":"","bonne_reponse":"A","explication":"cos(x + π\/4) = √2\/2 ⇒ x + π\/4 = π\/4 + 2kπ ou x + π\/4 = 7π\/4 + 2kπ ⇒ x = 2kπ ou x = 3π\/2 + 2kπ. Dans [0, 2π], les solutions sont x = 0 et x = 3π\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":285,"question":"Quelle identité trigonométrique permet de simplifier cos(x + a) ?","option_a":"cos(x + a) = cos(x)cos(a) - sin(x)sin(a)","option_b":"cos(x + a) = cos(x) + cos(a)","option_c":"cos(x + a) = sin(x)sin(a) - cos(x)cos(a)","option_d":"cos(x + a) = cos(x)cos(a) + sin(x)sin(a)","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'identité fondamentale pour cos(x + a) est cos(x + a) = cos(x)cos(a) - sin(x)sin(a), dérivée de la formule d'addition des angles.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":286,"question":"Si α = -1, combien de solutions l'équation cos(x + a) = -1 admet-elle dans l'intervalle [0, 2π] ?","option_a":"1 solution","option_b":"2 solutions","option_c":"Aucune solution","option_d":"4 solutions","option_e":"","option_f":"","bonne_reponse":"A","explication":"cos(x + a) = -1 admet une seule solution dans [0, 2π] : x + a = π + 2kπ ⇒ x = π - a + 2kπ. Dans [0, 2π], il n'y a qu'une seule valeur possible pour k.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":287,"question":"Quelle est la période des solutions de l'équation cos(x + a) = α ?","option_a":"2π","option_b":"π","option_c":"π\/2","option_d":"4π","option_e":"","option_f":"","bonne_reponse":"A","explication":"Les solutions de cos(x + a) = α sont de la forme x = ±arccos(α) - a + 2kπ, k ∈ ℤ. La période est donc 2π, car les fonctions trigonométriques sont périodiques de période 2π.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]