Question

已知函数f(x)=2根号3sinxcosx＋2cos平方x，求函数f(x)的最小正周期

要过程 谢谢！！！

Answer 1

f(x)=2（根号3）sinxcosx＋2cos平方x=（根号3）2sinxcosx+（2cos�0�5x-1）+1 =（根号3）sin2x+cos2x-1 =2【（�0�5×（根号3）×sin2x+�0�5×cos2x】+1 =2sin（2x+π/3）+1所以它的周期是T=2π/2=π

Answer 2

f(x)=2√3sinxcosx+2cos�0�5x=√3sin2x+cos2x+1=2(√3/2sinx+1/2cos2x)+1=2sin(2x+π/6)+1∴T=2π/2=π