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Find the maximum value of F (x) = (√ 3) SiNx + cos2x
F (x) = √ 3sinx + cos2x = √ 3sinx + (1-2sin & # 178; x) = - 2Sin & # 178; X + √ 3sinx + 1 = - 2 (SiNx - √ 3 / 4) & # 178; + 1 + 3 / 8 = - 2 (SiNx - √ 3 / 4) & # 178; + 11 / 8-1 ≤ SiNx ≤ 1, when SiNx = √ 3 / 4, the maximum value is 11 / 8; when SiNx = - 1, the minimum value is - 2 * (- 1 - √ 3 / 4) & # 178
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