Blue Hat(Mũ Xanh)
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Bảo | Chat Online Report 1/ 2√3.sin(x-π/8)cos(x-π/8) + 2cos²(x-π/8) = √3 + 4[sin²x + cos(π/3-x).cos(π/3+x)] ⇔ √3.sin(2x - π/4) + 1 + cos(2x - π/4) = √3 + 2(1 - cos2x) - 1 + 2cos2x ⇔ √3.sin(2x - π/4) + cos(2x - π/4) = √3 ⇔ √3/2.sin(2x - π/4) + 1/2.cos(2x - π/4) = √3/2 ⇔ sin(2x - π/4).cos(π/6) + cos(2x - π/4).sin(π/6) = √3/2 ⇔ sin(2x - π/4 + π/6) = √3/2 ⇔ sin(2x - π/12) = sin(π/3) ⇔ [ 2x - π/12 = π/3 + k2π ⇔[ x = 5π/24 + kπ (k ∈ Z) [ 2x - π/12 = 2π/3 + k2π [ x = 3π/8 + kπ (k ∈ Z) Vậy pt đã cho có các họ nghiệm là: ...
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