已知函数y=sin∧4x+2∫3sinxcosx-cos∧4x求该函数的最小值正周期和最小值

问题描述:

已知函数y=sin∧4x+2∫3sinxcosx-cos∧4x求该函数的最小值正周期和最小值

y=sin∧4x+2√3sinxcosx-cos∧4x (3前面是根号,不是积分?)=sin∧4x-cos∧4x+√3sin2x= (sin²x-cos²x)(sin²x+cos²x) + √3sin2x=√3sin2x - cos2x =2*sin(2x-π/6)周期T=2π/w=2π/2=...还有x属于【0,派】,求该函数的单调递增区间当x∈[0,π]∴2x-π/6∈ [ - π/6,11π/6]所以单调递增区间[- π/6, π/2] ∪ [ 3π/2,11π/6]画图可知谢谢了答题不易,要是认可,还请采纳,点击 满意 按钮,谢谢!!!