Given the function f (x) = 2sinx * sin (π / 2 + x) - 2Sin ^ 2x + 1. (2) if f (x0 / 2) = √ 2 / 3, x0 ∈ (- π / 4, π, 4), find the value of cos2x0?

Given the function f (x) = 2sinx * sin (π / 2 + x) - 2Sin ^ 2x + 1. (2) if f (x0 / 2) = √ 2 / 3, x0 ∈ (- π / 4, π, 4), find the value of cos2x0?

F (x) = 2sinx * sin (π / 2 + x) - 2Sin ^ 2x + 1 = 2sinxcosx + cos2x = sin2x + cos2x = √ 2Sin (2x + π / 4) because f (x0 / 2) = root 2 / 3, sin (x0 + π / 4) = 1 / 3cos2 (x0 + π / 4) = 1-2sin ² (x0+π/4)=1-2 × (1/3) ²= 7 / 9 is sin2x0 = - 7 / 9 and x0 ∈ (- π / 4, π

Given the function f (x) = 2Sin (2x + π / 6), if the function f (x0) = 2, find the value set composed of all x0 satisfying the condition

f(x0)=2sin(2x0+π/6)=2
sin(2x0+π/6)=1
2x0+π/6=π/2+2kπ,k∈Z
∴x0=π/6+kπ,k∈Z
{x0|x0=π/6+kπ,k∈Z}

Given the function f (x) = 2Sin (2X-4 / π), X ∈ R, if the function f (x) takes the maximum value at x = x0, find the value of F (x0) + F (2x0) + F (3x0)

2x-4/π=π/2+nπ（n=0.1.2.3……）
You can find the value of x0, and then bring it in

It is known that 3 = x + y under 8 + root sign, where x is an integer and 0 < y < 1, find the power of 2x + (3-y under root sign) 2009 two thousand and nine Find 2x + (3-y under the root sign)

∵1

It is known that x is equal to the 5th power of X + the 4th power of X - 10 times the 3rd power of X - 10 times the 2nd power of X + 2x + 1

x=√3-√2
x^5+x^4-10x^3-10x^2+2x+1
=x^2(x^2-10)(x+1)+2x+1
=(5-2√6)(-5-2√6)(√3-√2+1)+2(√3-√2)+1
=√3-√2

If y = root sign 1-2x + root sign 2x-1 + 2, find the value of the Y power of X

1/4; If the in the root sign should be greater than or equal to zero, then find x = 1 / 2, so y = 2