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f(x)=-10(√3)sinxcosx+10cos²x=-5(√3)sin2x+5(1+cos2x)=5[cos2x-(√3)sin2x]+5
=10cos(2x+π/3)+5;
∴①。最小正周期T=π; 单调增区间:[kπ-(2/3)π,kπ-(π/6)](k∈Z);
②。向右平移 π/6得g(x)=10cos(2x+π/6)+5;故由g(x)=10cos(2x+π/6)+5≧0,
得cos(2x+π/6)≧-1/2;故得2kπ-4π/3≦ 2x+π/6≦2kπ+2π/3,即kπ-3π/4≦x≦kπ+π/4;
=10cos(2x+π/3)+5;
∴①。最小正周期T=π; 单调增区间:[kπ-(2/3)π,kπ-(π/6)](k∈Z);
②。向右平移 π/6得g(x)=10cos(2x+π/6)+5;故由g(x)=10cos(2x+π/6)+5≧0,
得cos(2x+π/6)≧-1/2;故得2kπ-4π/3≦ 2x+π/6≦2kπ+2π/3,即kπ-3π/4≦x≦kπ+π/4;
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