1

On =( 1, √3)

Therefore: h(x)=sinx+√3cosx=2sin(x+π/3).

X∈[0, π/2], so: x+π/3∈[π/3, 5π/6]

Therefore: sin(x+π/3)∈[ 1/2, 1]

H(x)-t=0 always has two equal real roots on x∈[0, π/2].

That is, sin(x+π/3)=t/2 always has two equal real roots on x∈[0, π/2].

Therefore: √ 3/2 ≤ t/2.

Namely: √ 3 ≤ t

2

I want to know why1+cos (b+c) cos (b-c) =1-1/2cos (b-c)? Please explain.

A=π/3，A+B+C=π

Therefore: B+C=2π/3.

Therefore: 1+cos(B+C)cos(B-C)

= 1+cos(2π/3)cos(B-C)

= 1-cos(B-C)/2