∫tan 2θ=-2√2，

∴2tanθ/( 1-tan？ θ)=-2√2

∴tanθ=√2tan？ θ-√2

∴√2tan？ θ-tanθ-√2=0

∴tanθ=√2, or tanθ=-√2/2

∵2θ∈(π/2,π),∴θ∈(π/4,π/2)

∴tanθ>； 0 ∴tanθ=√2

∴sinθ=√6/3,cosθ=√3/3

∴(2cos^2(θ/2)-sinθ- 1)/[√2sin(π/4+θ)]

=(2cos？ θ-sinθ- 1)/(sinθ+cosθ)

=(2/3-√6/3- 1)/(√6/3+√3/3)

=-(6-3√2+√6-√3)/3

22. Let the function f(x) = ax 3-b/2x 2+c, and its image passes through (0, 1). (1) When the two roots of the equation f '(x)-x+ 1=0 are 65438+ respectively. (2) When a = 2/3 and b ≠ 0, find the maximum and minimum values of the function f(x).

(1)∫f(x) image intersection (0, 1). ∴c= 1.

'(x)=3ax^2-bx，

The two roots of equation f '(x)-x+ 1=0, namely 3ax 2-(b+ 1) x+ 1 = 0 are 1/2 and 1 respectively.

∴ (b+ 1)/(3a)=3/2， 1/(3a)= 1/2

∴a=2/3,b=2

∴f(x)=2/3x^3-x^2+ 1

(2)

f(x)=2/3x^3-b/2x^2+ 1

f'(x)=2x^2-bx=2x(x-b/2)

When b> is 0,

x & lt0，f '(x)& gt； 0, f(x) increasing,

0 & ltx & ltb/2，f '(x)& lt； 0, f(x) decreases.

X & gtb/2, f'(x)>0, and f(x) is increasing.

F (x) max =f(0)= 1

F(x) min = f (b/2) =-b 3/24+ 1。

When b< is 0,

x & ltB/2，f '(x)>； 0, f(x) increasing,

b/2 & lt； x & lt0，f '(x)& lt； 0, f(x) decreases.

x & gt0，f '(x)& gt； 0, f(x) increases.

∴f(x) Minimum =f(0)= 1

F(x) max = f (b/2) =-b 3/24+ 1。

23. if the function f(x) = ax 3-bx+4, when x=2, the function f(x) has an extreme value of -4/3.

(1) Find the analytical formula of function f(x); (2) If f(x)=k has three solutions, find the range of the number k.

f'(x)=3ax^2-b

When x=2, the function f(x) has an extreme value of -4/3.

∴f'(2)=0,f(2)=-4/3

∴ 12a-b=0

8a-2b+4=-4/3

∴a= 1/3,b=4

∴f(x)= 1/3x^3-4x+4

(2)

f'(x)=x^2-4=(x+2)(x-2)

∴f'(x)=0，x 1=-2，x2=2

∴x<； -2, increasing f(x),-2; 2, f(x) is increasing.

∴f(x) max =f(-2)=28/3，f(x) min =f(2)=-4/3。

If f(x)=k has three solutions, then -4/3.