Function y = asin (ω x+φ) (a >; 0，ω& gt； 0, | φ | less than π\2) In the same period, when x=π\4, y takes the maximum value of 2, and when x=7π\ 12, y takes the minimum value of -2.

The maximum value =2 and the minimum value =-2, so A=2.

Then the half period T/2=7π/ 12-π/4=π/3.

T=2π/3？ So what? w=3

f(x)=2sin(3x+φ)

Alternative point? (π/4,2)

then what sin(3π/4+φ)= 1？ 3π/4+φ=π/2 ? So φ=-π/4

So what?

( 1)f(x)=2sin(3x-π/4)

(2) If x ∈ [0,2π],

And f(x)=√3,

sin(3x-π/4)=√3/2

3x-π/4=2kπ+π/3 or 3x-π/4=2kπ+2π/3.

X=2kπ/3+7π/36 or x=2kπ/3+ 1 1π/36? k∈Z

2kπ/3+7π/36∈[0，2π]？ Or x = 2kπ/3+11π/36 ∈ [0,2π].

24kπ/36+7π/36x∈[0，2π]？ 24kπ/36+ 1 1π/36∈[0，2π]？

k=0？ x0=7π/36？ k=0？ x3= 1 1π/36

k = 1 x 1 = 3 1π/36k = 1？ x4=35π/36

(3) If the function f(x) satisfies the equation f (x) = a (0 < a < 2,), find the sum of all real number roots in [0,2π].

Can you tell by painting? The equation f (x) = a (0 < a < 2) has six roots in [0,2π].

x 1+x2=2*π/4=π/2

x3+x4=2*(π/4+2π/3)=π/2+4π/3

x5+x6=2*(π/4+4π/3)=π/2+8π/3

So the sum of all real numbers =3π/2+4π= 1 1π/4.