sin(2π/3-x)+cos²(π/6-x)=sin[π-(2π/3-x)]+sin²[π/2-（π/6-x）]
=sin(x+π/3)+sin²(x+3/π)
=1/4+(1/4)²
=5/16

有不明白的地方再问哟,祝你学习进步,更上一层楼! (*^__^*)