由第（1）题得,
当 x ∈[0,π]时,F(x) = sinx +sin（π/2 - x）= (√2)*cos(x- π/4)
同理,当x ∈(π,2π]时,
F(x) = - sinx +sin（π/2 - x）= 2cos[(π/2 -x + x) / 2 ] * sin[(π/2- x - x) / 2 ]
= (√2)*sin(π/4 - x)
= - (√2)*sin(x-π/4)
∵ x ∈[0,π]时,x-π/4 ∈[ - π/4,3π/4]
根据余弦函数性质,
当x-π/4 ∈[ - π/4,0),即x ∈[ 0,π/4) 时,F(x)=(√2)*cos(x- π/4) 单调递增,
当x-π/4 ∈[ 0,3π/4] ,即x ∈[ π/4,π] 时,F(x)=(√2)*cos(x- π/4) 单调递减.
而,当x ∈(π,2π]时,x-π/4 ∈ ( 3π/4,7π/4]
根据正弦函数性质,
当x-π/4 ∈( 3π/4,3π/2),即x ∈( π,7π/4)时,sin(x- π/4) 单调递减,则F(x)= - (√2)*sin(x-π/4)单调递增；
当x-π/4 ∈[ 3π/2,7π/4],即x ∈[ 7π/4,2π] 时,sin(x- π/4) 单调递增,则F(x)= - (√2)*sin(x-π/4)单调递减.
综上所述,当 x∈[ 0,π/4) ∪( π,7π/4)时,原函数F(x)单调递增,
当 x∈[ π/4,π] ∪[ 7π/4,2π] 时,原函数F(x)单调递减.