(1)
f(x)=cos²x
g(x)=1+(1/2)sin2x
f(x)=g(x)
(cosx)^2 =1+(1/2)sin2x
(cos2x +1)/2 = 1+(1/2)sin2x
sin2x- cos2x + 1 =0
√2sin(2x-π/4) =-1
2x-π/4 = -π/4
x = 0
ie a=0
(2)
y=f(x)
= (cosx)^2
(cos(x+x0))^2 = (cos(x-x0))^2
put x=x0
(cos(2x0))^2 = 1
2x0 = kπ
g(2x0) = 1+ (1/2)sin(kπ)
=1
(3)
h(x) =f(x) +g(x)
= (cosx)^2+ 1+(1/2)sin2x
= (1/2 )cos2x+(1/2)sin2x + 3/2
= (√2/2)(sin2x+π/4) + 3/2
h(x) :x∈[0,π/4]的值域
=[3/2,3/2+√2/2]