f(x)=6cos^2(wx/2)+√3sinwx-3=3( 1+coswx)+√3sinwx-3=√3sinwx+3coswx

f(x)=√3 sinwx+3c oswx = 2√3[sinwxcos(π/3)+coswxsin(π/3)]=

F(x)= 2√3 inches (wx+π/3)

Let AA' ⊥OX pass through point A in X' AA '=2√3. Because δδABC is a regular triangle, AA '=√3BA '=2√3, so BA '=2.

BA ' =( 1/4)* T = 2 = = & gt； T=8=2π/w == >w=π/4

f(x)=2√3sin[(π/4)x+π/3]

2)

2√3 sin[(π/4)x0+π/3]= 8√3/5 = = & gt； sin[(π/4)x0+π/3]=4/5 ①

- 10/3 & lt； x0 & lt2/3 == >-π/2 & lt； (π/4)x0+π/3 & lt； π/2

From ① we know that cos[(π/4)x0+π/3]=3/5 ②.

f(x0+ 1)= 2√3 sin[(π/4)(x0+ 1)+π/3]= 2√3 sin {[(π/4)x0+π/3]+π/4 } =

= sin[(π/4)x0+π/3]cos(π/4)+cos[(π/4)x0+π/3]sin(π/4)= 1

=(4/5)(√2/2)+(3/5)(√2/2)=7√2/ 10

So: f(x0+ 1)=7√2/ 10.

It's not the teacher's fault to change it to cos.