①θ=π/3 tanθ=tan(π/3)=√3
f(x)=sin²x+3cosx+3/8-3/2=1-cos²x+3cosx-9/8=-(cosx-3/2)²+9/4-1/8
f(x)最大值為-(1-3/2)²+9/4-1/8=-1/4+9/4-1/8=15/8
f(x)取最大值時需cosx=1 即 x=0
②f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2
=-cos²x+√3tanθcosx+﹙√3/8﹚tanθ-1/2
=-（cosx-√3tanθ/2)²+3tan²θ/4+﹙√3/8﹚tanθ-1/2
若θ∈[0,arctan(2/√3)] √3tanθ/2∈[0,1]
則f(x)的最大值為 3tan²θ/4+﹙√3/8﹚tanθ-1/2
若3tan²θ/4+﹙√3/8﹚tanθ-1/2=-1/8 解得tanθ=-√3/2 或tanθ=√3/3
因tanθ∈[0,(2/√3)] 所以 θ=π/6
若θ∈[arctan(2/√3),π/3] √3tanθ/2∈[1,√3]
則f(x)的最大值為 -（1-√3tanθ/2)²+3tan²θ/4+﹙√3/8﹚tanθ-1/2=9√3/8 tanθ-3/2
若9√3/8 tanθ-3/2=-1/8 得tanθ= 11√3/27 ,但 θ∈[arctan(2/√3),π/3]時 tanθ∈[2/√3,√3]
而 11√3/27