Let the maximum value of the function y = - 2Sin squared x-2acosx-2a + 1 on X be f (a). Try to determine that f (a) = 1 / 2 and find the minimum value of Y for the value of a, and the set of X when the minimum value is obtained

Let the maximum value of the function y = - 2Sin squared x-2acosx-2a + 1 on X be f (a). Try to determine that f (a) = 1 / 2 and find the minimum value of Y for the value of a, and the set of X when the minimum value is obtained

f(x)=－2sin²x－2acosx－2a＋1
f(x)=2cos²x－2acos－2a－1
f(x)=2×[cosx－(a/2)]²－[(1/2)a²＋2a＋1]
If the minimum value of function f (x) is f (a), then
{ f(－1)=1 (a2)
If f (a) = 1 / 2, then:
(1) If - 2 ≤ a ≤ 2,
Then: - (1 / 2) a 2-2a-1 = 1 / 2,
A = - 1
(2) If a > 2, then: 1-4a = 1 / 2,
Conclusion: a = 1 / 8
Thus: a = - 1, where f (x) = 2cos? X + 2cosx + 1 = 2 × [cosx + (1 / 2)] 2 + (1 / 2)
When cosx = - 1 / 2, y has a minimum value = 0 + 1 / 2 = 1 / 2
cosx=-1/2
x=2π/3+2kπ,k∈Z
Or x = 4 π / 3 + 2K π, K ∈ Z

Let the minimum value of function f (x) = cos2x-2acosx-2! Be - 7 and find the value of A

F (x) = 2cos? X-1-2acosx-2 * 1 = 2 (COS? X-acosx + (A / 2) 2 - (A / 2) 2) - 2 = 2 (cosx-a / 2) - a? 2 / 2-2 when 2 (cosx-a / 2) 2 = 0, the minimum value of F (x) is - a? / 2-2 = - 7, a = ±? 10

The minimum and maximum values of the function f (x) = cos2x + 2sinx are () A. -3，1 B. -2，2 C. -3，3 Two D. -2，3 Two

∵f(x)＝1−2sin2x+2sinx＝−2(sinx−1
2)2+3
2，
When SiNx = 1
2, Fmax (x) = 3
2，
When SiNx = - 1, Fmin (x) = - 3
Therefore, C

Let f [x] = - cos 2 x-4tsinx / 2cosx / 2 + 2T? - 6T + 2x ∈ R, the minimum value of F [x] is recorded as G [t] 〔1〕 Finding the expression of G [t] 〔2〕 When - 1 ≤ t ≤ 1, let the equation G [t] = KT have and only one real root

(1) f(x)=(sinx-t)^2+t^2
Therefore, G (T) = T ^ 2-6t + 1 (- 1 ≤ t ≤ 1)
(1-t)^2+t^2-6t+1=2t^2-8t+2(t＞1）
（-1-t）^2+t^2-6t+1=2t^2-4t+2(t＜-1）
(2) The first case is △ = [- (6 + k)] ^ 2-4 * 1 * 1 = 0 and K = - 4 or - 8
The second case is: let H (T) = T ^ 2 - (6 + k) t + 1
h(-1)*h(1)

Known vector A = (cosx, 2cosx), vector B = (2cosx, sin (π - x)), if f (x)= a• b+1． (1) Find the minimum period of the positive function (x); (2) If x ∈ [0, π 2] Find the maximum and minimum values of F (x)

（I）∵
a=(cosx，2cosx)，
b=(2cosx，sin(π-x))
∴f（x）=
a•
b+1=2cos2x+2cosxsin（π-x）+1
=1+cos2x+2sinxcosx+1
=cos2x+sin2x+2
=
2sin(2x+π
4)+2．
The minimum positive period of function f (x) t = 2 π
2=π．
（II）∵x∈[0，π
2]，
∴2x+π
4∈[π
4，5π
4]．
ν when + π
4=π
2, that is, x = π
At 8, f (x) has a maximum value of 2+
2；
When 2x + π
4=5π
4, that is, x = π
The minimum value of F (x) is 1

Function f (x) = cosx-2 cosx.sin ^2a/2-sinxsina(0

f(x)=cosx-2 cosx.sin ^2a/2-sinxsina=cosx(1-2sin^2a/2)-sinxsina=cosxcosa-sinxsina=cos(x+a)
Because x = π / 2 has a minimum value, and 0

Given the function f (x) = 2cosx + sin square x, find the maximum value of F (x)

f(x)=2cosx+sin^2 x
=-cos^2 x+2cosx+1
Let t = cosx
Then f (x) = - T ^ 2 + 2T + 1 = - (t-1) ^ 2 + 2
Because t ∈ [- 1,1]
So when t = 1, f (x) has a maximum value of 2

When π / 6

The function y = 3-sinx-2cosx ^ 2 is deformed because 1 = SiNx ^ 2 + cosx ^ 2, so the original formula = 1 + 2-sinx-2cosx ^ 2 = 1 + 2 (SiNx ^ 2 + cosx ^ 2) - sinx-2cosx ^ 2 = 1 + 2sinx ^ 2-sinx = 2 (sinx-1 / 4) ^ 2 + 7 / 8 because of π / 6

Find the minimum and maximum values of the function y = sinx-2cosx (0 ≤ x ≤ π), and find the corresponding angle X

y=√5sin(x-z)
Where Tanz = 2
Because Tan π / 4 = 1
So π / 4

Find the maximum, minimum and minimum positive period of the function y = 2 √ 3 sinx-2cosx + 6

y=2√3 sinx-2cosx+6
=4sin(x-π/6)+6
The maximum value is 10 and the minimum value is 2
Minimum positive period T = 2 π