已知函数f（x）=2cos的平方（x+π/12）+sin2x 1若f(a)=1 a∈(0,π)求a的值 2求f(x)的单调增区间

1、f(x)=2[1+cos(2x+π/6)]/2+sin2x=cos2xcosπ/6-sin2xsinπ/6+sin2x+1=cos2xcosπ/6-1/2*sin2x+sin2x+1=cos2xcosπ/6+1/2*sin2x+1=cos2xcosπ/6+sin2xsinπ/6+1=cos(2x-π/6)+1f(a)=1cos(2a-π/6)=02a-π/6=kπ+π...