求函数f（X）=1/cos2x+2/sin2x(x为锐角)的最小值要 过程啊谢谢
热心网友
f(x)= 1/cos^2(x) +2/sin^2(x) = [sin^2(x) + cos^2(x)]*[1/cos^2(x) +2/sin^2(x)] = sin^2(x)/cos^2(x) + 2 + 1 + 2cos^2(x)/sin^2(x) =tan^2(x) + 2cot^2(x) +3 ≥2√[tan^2(x)*2cot^2(x)] +3 =2√2 +3当且仅当tan^2(x) = 2cot^2(x)时取得最小值。注：遇到a/x + b/(1-x)的形式时，一般都采用乘以[x+(1-x)]去分母。