Find the set of X which makes the function y = 2sin2x obtain the maximum and minimum values, and point out the maximum and minimum values

Find the set of X which makes the function y = 2sin2x obtain the maximum and minimum values, and point out the maximum and minimum values

{X|X=Kπ/2+π/4,K∈Z}
The maximum value is 2 and the minimum value is - 2

Given that the function y = 2sin2x + csc2x + TaNx + Cotx, X belongs to (0,90 °), to find the minimum value of Y, there is the following solution 2sin2x + csc2x = 2sin2x + 1 / sin2x > = 2 radical 2, TaNx + 1 / TaNx > = 2, the sum of the two formulas yields Y > = 2 radical 2 + 2, so Ymin = 2 radical sign 2 + 2 1. Try to judge whether the above solution is correct 2. If you think the above solution is correct, please write down the value of independent variable x when y gets the minimum value; if you think the above solution is not correct, please explain the reason and find the minimum value of function y

1. Not true
∵ 2sin2x + csc2x = 2sin2x + 1 / sin2x > = 2 root sign 2. When SiNx = 1, get "="
TaNx + 1 / TaNx > = 2 when TaNx = 1, get "="
Cannot get "=" at the same time
y＝2sin2x＋csc2x＋tanx＋cotx＝2sin2x＋1/sin2x＋sinx/cosx＋cosx/sinx
＝2sin2x＋1/sin2x＋1/(sincosx)＝2sin2x＋3/sin2x
＝sin2x＋sin2x＋3/(4sin2x)＋3/(4sin2x)＋3/(4sin2x)＋3/(4sin2x)
≥6[sin2x×sin2x×3/(4sin2x)×3/(4sin2x)×3/(4sin2x)×3/(4sin2x)]^(1/6)＝6׳√(3/4)²
The minimum value of y = 3 √ 36 / 4
In this case, sin2x = 3 / (4sin2x) sin? 2x = 3 / 4 ∵ x ∈ (0,90 °)

Find the value range of the function f (x) = Tan ^ 2x + 2a, TaNx + 5 when x ∈ [π / 4, π / 2] (where a is a constant)

Is the symbol "Tan ^ 2x" wrong

Find the function f (x) = tan2x + 2atanx + 5, X ∈ [π 4，π 2) Where a is a constant

∵x∈[π
4，π
2) Let TaNx = t ≥ 1, then the function f (x) = H (T) = T2 + 2at + 5, and the axis of symmetry is t = - A,
When a ≥ - 1, - a ≤ 1, t = 1, the minimum value of function H (T) is 6 + 2a and the range of original function value is [6 + 2a, + ∞)
When a ＜ - 1, - a > 1, t = - A, the minimum value of function H (T) is - 5-a2, and the range of original function value is [5-a2, + ∞)

If x ∈ (0), \ \ frac {π} {2}) find the minimum value of function y = TaNx + Tan (\ \ frac {π} {2} - x)

Because x ∈ (0, PI / 2), Tan (x) ∈ (0, positive infinity)
y=tan(x)+tan(pi/2-x)=tan(x)+cot(x)>=2genhao(tan(x)*cot(x))=2
When x = pi / 4, the equal sign is taken

If x ∈ [- π / 3,2 π / 3], find the maximum and minimum of the function y = cos ^ 2 (x + π / 6) + sin (x + 2 π / 3)

The original formula can be reduced to y = sin 2 (π / 3-x) + sin (π / 3-x)
Let t = sin (π / 3-x)
Then y = t? 2 + T = (T + 1 / 2) 2 - 1 / 4
∵x∈[-π/3,2π/3]
∴π/3-x∈[-π/3,2π/3]
ν t ∈ [root of negative half sign 3,1]
So y ∈ [- 1 / 4,2]
The maximum value of Y is 2 and the minimum value is - 1 / 4

The known function f (x) = sin2 (x - π) 6）+cos2（x-π 3）+sinx•cosx，x∈R． (1) Find the maximum value of F (x) and the value of X when obtaining the maximum value; (2) Find the monotone increasing interval of F (x) on [0, π]

(1) If f (x) = (sinxcos π 6 − cosxsin π 6) 2 + 2 + (cosxcos π 3 + sinxsin π 3) 2 + SiNx * cosx = sin2x + SiNx + cosx + 12 = 12 (sin2x − cos2x) + 1 = 22sin (2x − π 4) + 1 = 1, when 2x − π 4 = π 2 + 2K π (K ∈ z), namely x = 3 π 8 + K π (K ∈ z), the function f (... Z) is the function f (... Z) when the function f (... Z) is the function f (... Z) of the function f (... Z), the function f (... Z), the function f (... Z), the function f (... Z), the function f (it's a good idea

The maximum value of the function y = radical 3 / 2 * sin (x + Pai / 2) + cos (Pie / 6-x)

Y = radical 3 / 2 * sin (x + Pai / 2) + cos (PAI / 6-x) = (√ 3 / 2) cosx + cos (π / 6) cosx + sin (π / 6) SiNx = √ 3cosx + (1 / 2) SiNx = √ (1 / 4 + 3) * sin (x + {0)} is an acute angle, and Tan} = √ 3 / (1 / 2) = 2 √ 3 = (√ 13 / 2) * sin (x + {x +)

The maximum value of three sin (x + Pai / 2) + cos (Pie / 6-x) is

=-Root of two thirds three cosx + (COS π / 6 cosx + sin π / 6 SiNx)
=-Half root three cosx + (half root three cosx + 1 / 2sinx)
=1/2 sinx
Because - 1

(tanx+cotx)cos2x

The original formula = (SiNx / cosx + cosx / SiNx) cos2x
=((sinx)^2+(cosx)^2)/(sinx*cosx)*cos2x
=1/(sinx*cosx)*cos2x
=2cos2x/sin2x
=2cot2x