18. Given the function f (x) = sin2x + 2cos ^ 2x-1, find the maximum and minimum values of the function f (x) on the interval [π / 4,3 π / 4]

18. Given the function f (x) = sin2x + 2cos ^ 2x-1, find the maximum and minimum values of the function f (x) on the interval [π / 4,3 π / 4]

f（x）
＝sin2x＋2（cosx）^2－1＝sin2x＋cos2x＝√2［sin2xcos（π/4）＋cos2xsin（π/4）］
＝√2sin（2x＋π/4）.
∵π/4≦x≦3π/4,∴π/2≦2x≦3π/2,∴π/2＋π/4≦2x＋π/4≦3π/2＋π/4.
The maximum value of F (x) = 2Sin (π / 2 + π / 4) = 2cos (π / 4) = 1
The minimum value of F (x) = 2Sin (3 π / 2) = - √ 2

Given that the function f (x) = cos2x + sin2x, X belongs to R. the minimum and maximum values of the function f (x) in the interval 〔 - π / 8, π / 2 〕 are obtained

If (x) = cos2x + sin2x = 2, 2 (sin2x) is the most important factor in the world, such as (sin2x) + (2 / 2) + (cos2x, 2 / 2), and (2 / 2), 2 / 2 (2 / 2), 2 / 2 (2 / 2), 2 [sin2xcos (π / 4) + cos2xsin (π / 4) + cos2xsin (π / 4) + cos2x (π / 4) + π / 8 ≤ x ≤ x ≤ π / 2 = = > 0 ≤ 2x + π / 4 ≤ 5 π / 4 - / 2 / 2 ≤ sin (2x + π / 4) ≤ 1 = = > - 1 ≤ f (x) ≤ 1 ≤ f (x) ≤ 1 ≤ 1 ≤ f (x) ≤ 1 ≤ 1 ≤ f (x) ≤ 1 ≤ 2F (max) = √ 2 when

Find the minimum value of the function y = sin2x + 2sinxcosx + 2 √ 3cos2x, and write the set of X that makes the function y get the minimum value Urgent! Online, etc

Because 2sinxcosx = sin2x
Then y = sin2x + 2sinxcosx + 2 √ 3cos2x
=sin2x+sin2x+2√3cos2x
= 2sin2x +2√3cos2x
=2（sin2x +√3cos2x）
=4（1/2 sin2x +√3/2 cos2x）
=4(sin2x cosπ/3 +cos2x sinπ/3)
=4sin(2x +π/3)
So the minimum value of function y is - 4
Because when 2x + π / 3 = 3 π / 2 + 2K π (k is an integer), the minimum value of function y is - 4,
That is, x = (3 π / 2 - π / 3) / 2 + K π,
x=7π/12+kπ ,
Therefore, the set of X with the minimum value of function is {x | x = 7 π / 12 + K π, and K is an integer}

Find the minimum value of the function y = sin2x + 2sinxcosx + 3cos2x, and write the set of X that makes the function y take the minimum value

Y = sin2x + 2sinxcosx + 3cos2x = (sin2x + cos2x) + 2sinxcosx + 2cos2x = 1 + sin2x + (1 + cos2x) = 2 + sin2x + cos2x = 2 + 2Sin (2x + π 4)

The maximum and minimum values of the function y = sin2x + root 3cos2x + 2 As the title

2sin（2x+π/3）+2
Maximum 4, minimum 0

If the minimum value of F (x) = 2cos 2 x-2a-1-2acosx is a function of a, it is noted that G (a) can find the expression of G (a) d (2) to find the value that can make g (a) = 1 / 2, Also find the maximum value of F (x) when a takes this value

1）y=2cos^2x-2acosx-(2a+1)
=2[(cosx-a/2)^2]-(a^2/2+2a+1)
When - 1 ≤ A / 2 ≤ 1, that is - 2 ≤ a ≤ 2, yman = - (a ^ 2 / 2 + 2A + 1)
When a / 2 > 1, that is, a > 2, Ymin = y|x = 1 = 2-2a - (2a + 1) = - 4A + 1
When a / 2

Let the minimum value of the function y = 2cos2x-2acosx - (2a + 1) be f (a), and try to make sure that f (a) = 1 2, and find the maximum value of Y for the value of a at this time

Let cosx = t, t ∈ [- 1,1],
Then y = 2t2-2at - (2a + 1), and the axis of symmetry t = a
2，
When a
2 ＜− 1, that is, when a ＜ - 2, [- 1,1] is the increasing interval of function y, and Ymin = 1 ≠ 1
2；
When a
When 2 > 1, i.e. a > 2, [- 1,1] is the decreasing interval of function y, and Ymin = - 4A + 1 = 1
2，
A = 1
8, and a ＞ 2;
When − 1 ≤ a
When 2 ≤ 1, i.e. - 2 ≤ a ≤ 2, Ymin = − A2
2−2a−1＝1
2，a2+4a+3＝0
A = - 1, or a = - 3,
∴a=-1，
In this case, ymax = - 4A + 1 = 5

Given the function y = 2cos ^ 2x-2acosx - (2a + 1), find the minimum value of function f (a)____ I figured out that the minimum is - 2a-1-a ^ 2 / 2 (I don't know if it's right) 201 (1) Try to determine the value of a satisfying f (a) = 1 / 2 (2) When a takes the value in (1), find the maximum value of Y,

Let t = cosx, then | t | < = 1
y(t)=2t^2-2at-(2a+1)=2(t-a/2)^2-(a^2/2+2a+1)
The opening is upward, and the axis of symmetry is x = A / 2
It is necessary to discuss the size of a to determine the minimum value
When - 2 = when a > 2, the symmetry axis is on the right side of the interval, and the minimum value is obtained when t = 1, f (a) = 1-4a
When a < - 2, the axis of symmetry is on the left and left of the interval. When t = - 1, the minimum value is obtained, f (a) = 1
1) If a > 2, then f (a) = 1-4a = 1 / 2, then a = 1 / 8
If a < - 2, f (a) = 1 also does not match
If - 2 = then a = - 1
2)a=-1, y=2(t+1/2)^2+1/2
When t = 1, y takes the maximum value ymax = 5

The minimum value of the function f (x) = 1-2a-2acosx-2sin ^ 2 is g (a), a ∈ R (1.) (2) If G (a) = 1 / 2, find the maximum value of a and f (x)

1) Let t = cosx
Then f (x) = 1-2a-2at-2 (1-T ^ 2) = 2T ^ 2-2at-2a-1 = 2 (T-A / 2) ^ 2-A ^ 2 / 2-2a-1
Because | t | < = 1, it is necessary to divide the symmetry axis X = A / 2 into three sections to discuss g (a)
When a / 2 < - 1, the minimum is t = - 1, and G (a) = 1
When - 1 = when a / 2 > 1, when t = 1, G (a) = 1-4a
2) If a / 2 > 1, then G (a) = 1-4a = 1 / 2, then a = 1 / 8, contradiction;
If - 1 = only a = - 1
In this case, f (x) = 2 (T + 1 / 2) ^ 2 + 1 / 2
When t = 1, the maximum value is 5

Given 0 ≤ x ≤ π / 2, find the minimum value of the function y = - (SiNx) ^ 2-2acosx + 1

y=-(sinx)^2-2acosx+1
=cos²x-2acosx
=(cosx-a)²-a²
If 1 = > a > = 0, the minimum value is obtained at x = arccosa, and the minimum value is y = - a 2
If a > 1, the minimum value is obtained when x = 0, and the minimum value is y = (1-A) 2 - a 2 = 1-2a
If a