Given the function f (x) = 1 - √ 3sin (2x) + 2cos & # 178; (x), let the opposite sides of angles a, B and C of triangle ABC be a, B and C respectively, and a = 1, F (a) = 0, find the value range of B + C
From F (a) = 0, 1 - √ 3sin (2a) + 2cos & # 178; (a) = 0sin & # 178; (a) + cos & # 178; (a) - 2 √ 3sin (a) cos (a) + 2cos & # 178; (a) = 0 Sin & # 178; (a) - 2 √ 3sin (a) cos (a) + 3cos & # 178; (a) = 0
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