It is known that a, B and C are the opposite sides of three internal angles a, B and C of triangle ABC, 2bcosc = 2a-c If the area of triangle ABC is root 3, find the value range of B. (possible values, I calculate B = π / 3

It is known that a, B and C are the opposite sides of three internal angles a, B and C of triangle ABC, 2bcosc = 2a-c If the area of triangle ABC is root 3, find the value range of B. (possible values, I calculate B = π / 3

(1) from the sine theorem: 2sinbcosc = 2sina sinc,
In △ ABC, Sina = sin (B + C) = sinbcosc + cosbsinc,
∴2cosBsinC=sinC,
∵ C is the inner angle of a triangle, so we can get sinc > 0
∴cosB=1/2,
∵ B is the inner angle of a triangle, B ∈ (0, π),
∴B=π/3；
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⑵S△ABC=1/2×acsinB=√3,
B=π/3,sinB=√3/2,
∴ac=4,
From the cosine theorem, we get: B & # 178; = A & # 178; + C & # 178; - 2Ac & # 9642; CoSb = A & # 178; + C & # 178; - AC ≥ 2ac-ac = AC = 4 (if and only if a = C = 2, the equal sign holds)
[(A-C) ² ≥ 0 → A & #178; + C & #178; ≥ 2Ac if and only if a = C]
So the value range of B is [2, + ∞)
Sine theorem; cosine theorem
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