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f(x)=2sinx(√3/2sinx+1/2cosx)=√3(sinx)^2+sinxcosx=√3/2(1-cos2x)+1/2sin2x
=√3/2+(sin2xcosπ/3-cos2xsinπ/3)=√3/2+sin(2x-π/3)
=√3/2+(sin2xcosπ/3-cos2xsinπ/3)=√3/2+sin(2x-π/3)
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f(x)
=2sinx.[(√3/2)sinx+ (1/2)cosx ]
=√3(sinx)^2 + sinx.cosx
= (√3/2) .( 1-cos2x) + (1/2)sin2x
=sin(2x-π/3) +√3/2
=2sinx.[(√3/2)sinx+ (1/2)cosx ]
=√3(sinx)^2 + sinx.cosx
= (√3/2) .( 1-cos2x) + (1/2)sin2x
=sin(2x-π/3) +√3/2
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