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设锐角三角形ABC中,2sin^2A-cos^2A=2.(1)求

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设锐角三角形ABC中,2sin^2A-cos^2A=2.(1)求
2sin^2A-cos^2A=2
3sin^2A-(sin^2A+cos^2A)=2
3sin^2A=2+(sin^2A+cos^2A)=2+1=3
sinA=1
(1),∠A=90°
(2),
y=2sin^2B+sin(2B+π/6)
=1-cos(2B)+(0.5√3)*sin(2B)+0.5cos(2B)
=1+(0.5√3)*sin(2B)-0.5cos(2B)
=1+sin(2B-π/6)
=取最大值时,sin(2B-π/6)=1
2B-π/6=π/2
∠B=π/3