设x≤a≤π,不等式8 x2-(8sina)x cos2a≥0对x∈R恒成立,则a的取值范围.
问题描述:
设x≤a≤π,不等式8 x2-(8sina)x cos2a≥0对x∈R恒成立,则a的取值范围.
答
你好!应该是 0 ≤ a ≤ π8x² - (8sina) x + cos2a ≥ 0 恒成立∴ Δ = 64sin²a - 32cos2a ≤ 02sin²a - cos2a ≤ 01 - cos2a - cos2a ≤ 0 cos2a ≥ 1/2 0 ≤ a ≤ π0 ≤ 2a ≤ 2π∴ 0 ≤ 2a ≤ ...