20 pts...Bjr à tous j'espère que vous allez bien.Pourriez vous m'aider merci d'avance.... Pergunta de ideia de1D2010

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May 2019 1 126 Report

20 pts...Bjr à tous j'espère que vous allez bien.Pourriez vous m'aider merci d'avance

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scoladan

Verified answer

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

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