Vector M = (2sinx, 1), n = (√ 3cosx, 2cos2x), function f (x) = mn-t. (1) if the equation f (x) = 0 has a solution on 0 < = x < = π / 2, find the range of T value; (2) In the triangle ABC, ABC is the opposite side of ABC respectively. When t in (1) is the maximum value and f (a) = - 1, B + C = 2, the minimum value of a is obtained

Vector M = (2sinx, 1), n = (√ 3cosx, 2cos2x), function f (x) = mn-t. (1) if the equation f (x) = 0 has a solution on 0 < = x < = π / 2, find the range of T value; (2) In the triangle ABC, ABC is the opposite side of ABC respectively. When t in (1) is the maximum value and f (a) = - 1, B + C = 2, the minimum value of a is obtained

1:f(x)=2√3sinxcosx+2cos2x-t=√3sin2x+2con2x-t=2sin(2x+π/6)-t
f(x)=0,2sin(2x+π/6)＝t,
Zero

Given 2cosx + cosy = 2 / 3, 2sinx siny = 3 / 4, find cos (x + y)

(2cosx+cosy)^2+(2sinx-siny)^2
=4(cosx)^2+4(sinx^2+(cosy)^2+(siny)^2+4(cosxcosy-sinxsiny)
=5+4cos(x+y)
=4/9+9/16
=145/144
cos(x+y)=-575/576

Cos (X-2 / 3 π) - cosx = (root 3) / 2sinx-3 / 2cosx = root 3sin (x - π / 3) How did these steps come about?

Cos (X-2 / 3 π) - cosx = cosxcos2 / 3 π + sinxsin2 / 3 π - cosx = - 1 / 2cosx + (root 3) / 2sinx cosx
=(Gen 3) / 2sinx-3 / 2cosx
=(root 3) * (sinxcos π / 3-sin π / 3cosx)
=(radical 3) sin (x - π / 3)
First decompose, then merge, and finally extract the root sign 3 to get the formula of sin (x - π / 3)

Small period of F (x) = cos squared x + sinxcosx + 3 / 2 x ∈ R (1) f (x) and f (π / 8) The small period of F (x) = cos square x + sinxcosx + 3 / 2 x ∈ R (1) f (x) and the value of F (π / 8) (2) find the monotone increasing interval of this function

(1 / 2) sin2x + 3 / 2 = (1 / 2) sin2x + 3 / 2 = (1 / 2) sin2x + (1 / 2) cos2x + 2 = (2 / 2) sin2x + (1 / 2) cos2x + 2 = (2 / 2 / 2) sin (2x + π / 4) + 2 (1) t = 2 π / 2 = π, f (π / 8) = (√ 2 / 2) + 2 (2) 2K π - π / 2 ≤ 2x + π / 4 ≤ 2K π + π + π / 2: K π - 3 π / 8 ≤ x ≤ K π + π / 8 / 8 ≤ K π + 8 / 8: K π - 3 π - 3 π - 8 ≤ x ≤ K π + the increasing range is: [K π - 3 π / 8, K π + π / 8]

The minimum positive period of the function y = cos2x is______ .

∵ function y = cos2x = 1 + cos2x
2=1
2cos2x+1
2，
The minimum positive period of the function y = cos2x is 2 π
2=π，
So the answer is: π

The minimum positive period of the function y = 2 (COS square x) is

PI (PIE)

Given the function f (x) = 2Sin (1 / 2x + π / 3) + 1, the maximum and minimum values of X ∈ R are 3 and - 1 respectively (3) If the function f = f (x) + LNK has and only two zeros on [- π / 6, π], the value of real number K

A:
f(x)=2sin(x/2+π/3)+1
The maximum value is 2 + 1 = 3, and the minimum value is - 2 + 1 = - 1
F (x) = f (x) + LNK = 2Sin (x / 2 + π / 3) + 1 + LNK = 0 has and only two zeros on [- π / 6, π]
So: - (1 + LNK) = 2Sin (x / 2 + π / 3)
Because: - π / 6 < = x < = π
So: - π / 12 < = x / 2 < = π / 2, - π / 12 + π / 3 < = x / 2 + π / 3 < = π / 2 + π / 3
So: π / 4 < = x / 2 + π / 3 < = 5 π / 6
So: sin (5 π / 6) < = sin (x / 2 + π / 3) < = sin (π / 2)
So: 1 / 2 < = sin (x / 2 + π / 3) < = 1
Therefore, 2 * sin (π / 4) < = - (1 + LNK) < 2 * 1
So: √ 2 < = - (1 + LNK) < 2
Therefore: - 2 < 1 + LNK < = - √ 2
Therefore: - 3 solution: 1 / e ^ 3 this problem needs to draw a sketch combined with understanding

Given the function f (x) = 2Sin (2x + π - 3), find the maximum and minimum values of F (x) on the interval [- 30 ° and 90 °]

Because x is on the interval [- 30 ° and 90 °], so (2x + π - 3) is in the interval [0 ° 240 °], the maximum value of F (x) is 2 and the minimum value is - √ 3

What are the minimum and maximum values of the function y = 1 / 2Sin (π / 4-2x / 3) on the interval [- π / 8,11 π / 8]

The main use of the image, the coefficient of X first into positive
y＝(1/2)sin(π/4－2x/3)=(-1/2)sin(2x/3-π/4)
Then, let t = 2x / 3-U / 4, take it as the whole angle and find its range
x∈[－π/8,11π/8]
2x/3-π/4∈[-π/3,2π/3]
Then draw the image of y = - 1 / 2sint, take the image of corresponding interval, and the maximum value will be clear at a glance, y (max) = y (- Wu / 3) = √ 3 / 4, y (min) = y (Wu / 2) = - 1 / 21| comment (1)

It is known that f (x) = 2Sin (2x - π / 3) + 1, X ∈ [π / 4, π / 2] If the inequality f (x) - M ＜ 2 holds on X ∈ [π / 4, π / 2], find the value range of real number M

f(x)=2sin(2x-π/3)+1,x∈[π/4,π/2]
∵ x∈[π/4,π/2]
∴ 2x-π/3∈[π/6,2π/3]
∴ sin(2x-π/3)∈[1/2,1]
∴ f(x)=2sin(2x-π/3)+1∈[2,3]
∵ |f(x)-m|