(1) If m→? N→=2, and find the value of cos (x+π 3);

(2) Remember that f(x)=m→? N→, in △ABC, the opposite sides of angle A. B. C are A, B and C respectively, and satisfy (2a? C)cosB=bcosC, and find the range of f(A).

Maximum value of trigonometric function, operation of plane vector product, cosine function of sum and difference of two angles.

( 1) sin(

x

2

+

Pi?

six

), and then find cos(x+

Pi?

three

) value.

(2) Use sine theorem to find the value of b through (2a-c)cosB=bcosC, find the value range of a through the sum of internal angles of a triangle, and then find the value range of f(A).

( 1)m→? n→= 23√sin x4 cos x4+2 cos 2x 4 = 3√sin x2+cos x2+ 1

=2sin(x2+π6)+ 1。

∵m→? n→=2

∴sin(x2+π6)= 12.

cos(x+π3)= 1？ 2sin2(x2+π6)= 12。

(2)∵(2a？ c)cosB=bcosC，

(2sinA？ sinC)cosB=sinBcosC，

∴2sinAcosB？ sinCcosB=sinBcosC，

∴2sinAcosB=sin(B+C).

∵A+B+C=π,∴sin(B+C)sinA, and Sina ≠0,

∴cosB= 12,B=π3，

∴0<； A & lt2π3.∴π6<； a2+π6 & lt； π2 12 & lt； sin(A2+π6)& lt； 1

∫f(x)= m→n→=2sin(x2+π6)+ 1,∴f(a)=2sin(a2+π6)+ 1

So the range of f(A) is (2,3).