1.f(x)=cos^2x-3cosx+sin^2x-3sinx
=-3√2sin(x+π/4)+1
x+π/4∈[2kπ+π/2,2kπ+3π/2]
因为x∈【-π,0】
所以x∈【-π,-3π/4】
2.=-3√2sin(x+π/4)+1=-1
sin(x+π/4)=√2/3
x∈【-π/4,π/4】
x+π/4∈【0,π/2】
所以cos(x+π/4)>0=√7/3
cos2x=sin(2x+π/2）=2sin(x+π/4)cos(x+π/4)=2√14/9
x∈【-π/4,π/4】
2x∈【-π/2,π/2】
所以sin2x=5/9or-5/9
所以tan2x=5√14/28or-5√14/28