f(x)=5sinxco0sx-5√3cos^2x+5/2√3
=1\2(10sinxcosx-10√3cos^2x+5√3 )
=5\2(sin2x-√3cos2x)
=5[(1\2)sin2x-(1\2)√3cos2x]
=5sin(2x-π\3)