(1)
f(x)=1-2a-2acosx-2sin²x
=-2(1-cos²x)-2acosx-2a+1
=2cos²x-2acosx-2a-1
=2(cosx-a/2)²-a²/4-2a-1-1≤cosx≤1
當(dāng)-1≤a/2≤1即-2≤a≤2時,
當(dāng)cosx=a/2時,g(a)=f(x)min=-a²/4-2a-1

當(dāng)a/2<-1即a<-2時,
當(dāng)cosx=-1時,g(a)=f(x)min=1


當(dāng)a/2>1即a>2時,

cosx=1時,g(a)=f(x)min=1-4a
綜上所述
{1,(a<-2)
g(a)={ -a²/2-2a-1(-2≤a≤2)
{ 1-4a, (a>2) (2)

g(a)=1/2

a<-2時,不存在a值

-2≤a≤2時,

由-a²/2-2a-1=1/2

得a²+4a+3=0,
a=-3（舍去）或a=-1
當(dāng)a>2時,由1-4a=1/2得：a=1/8,不合題意 滿足條件的a=-1
此時,f(x)=2(cosx+1/2)²+1/2
cosx=1時,f(x)max=5