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f(x)
=√3cos2x-2[cos(x+π/4)]^2 +1
=√3cos2x-{ 2[cos(x+π/4)]^2 -1 }
=√3cos2x-cos[2(x+π/4)]
=√3cos2x-cos(π/2 +2x)
=√3cos2x+ sin2x
=2sin(2x+π/3)
=√3cos2x-2[cos(x+π/4)]^2 +1
=√3cos2x-{ 2[cos(x+π/4)]^2 -1 }
=√3cos2x-cos[2(x+π/4)]
=√3cos2x-cos(π/2 +2x)
=√3cos2x+ sin2x
=2sin(2x+π/3)
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