题目

已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+4,已知f(α)=5,求tanα的值（要具体过程）
注：该化简结果是：f(x)=2sin(2x+π/6)+4,tanα=0或根号3（请问具体过程是什么?）

提问时间：2021-03-30

答案

f(x)=sin(2x+π/6）+sin2xcosπ/6+cos2xsinπ/6-cos2x+4
=sin(2x+π/6）+sin2xcosπ/6+cos2x*1/2-cos2x+4
=sin(2x+π/6）+sin2xcosπ/6-cos2x*1/2+4
=sin(2x+π/6)+sin2xcosπ/6-cos2xsinπ/6+4
=sin(2x+π/6)+sin(2x+π/6)+4
=2sin(2x+π/6)+4
所以2sin(2α+π/6)+4=5
sin(2α+π/6)=1/2
2α+π/6=2kπ+π/6,2α+π/6=2kπ+5π/6
α=kπ,α=kπ+π/3
tanα=0或根号3