Let f (x) = cos [2x + / 3] + sin2x, find the maximum and minimum positive period of F (x)

Let f (x) = cos [2x + / 3] + sin2x, find the maximum and minimum positive period of F (x)

First, expand the function to get cos2x. Multiply the root sign 3 by 2-1 / 2sin2x + sin2x, which is equivalent to cos2x. Multiply the root sign 3 by 2 + 3 / 2sin2x, and then transform it into an angular trigonometric function to get the root sign 3 times sin (2x + k). Then we can find that the maximum value is root sign 3 and the minimum positive period is Pai

It is known that the maximum value of the function f (x) = 2 radical 3sin (x + π / 4) cos (x + π / 4) + sin2x + A is 1 Help~
It is known that the maximum value of the function f (x) = 2 radical 3sin (x + π / 4) cos (x + π / 4) + sin2x + A is 1,
(1) Finding the value of constant a
(2) Finding monotone increasing interval of function f (x)
(3) If the image of F (x) is shifted to the left by π / 6 units, the image of function g (x) is obtained, and the maximum and minimum values of function g (x) in the interval [0, π / 2] are obtained
Make as much as you can --

F (x) = 2 √ 3sin (x + π / 4) cos (x + π / 4) + sin2x + a = √ 3sin (2x + π / 2) + sin2x + a = √ 3cos2x + sin2x + a = 2Sin (2x + π / 3) + a obviously: a + 2 = 1 a = - 1 the monotone increasing interval of function f (x): 2K π - π / 2 ≤ 2x + π / 3 ≤ 2K π + π / 2K - 5 π / 12 ≤ x ≤ K π + π / 12 is [K π - 5

The value of 3sina + cosa = 0,1 / cos square a + sin square a

I don't like sleeping with one woman many times, but I like sleeping with many women only once

Finding the period and range of y = SiNx + cosx

y=sinx+cosx=√ 2sin(x+π/4).
So the period T = 2 π; the range is: [- √ 2, √ 2]

How to find the range of SiNx / (2-cosx)

Y = SiNx / (2-cosx) y (2-cosx) = SiNx 2Y ycosx = SiNx SiNx + ycosx = 2Y. From the auxiliary angle formula of trigonometric function, we can see that | SiNx + ycosx ≤ √ (1 + Y & sup2;) so | 2Y ≤ √ (1 + Y & sup2;) 4Y & sup2; ≤ 1 + Y & sup2; 3Y & sup2; ≤ 1 - √ 3 / 3 ≤ y ≤ √ 3 / 3 function range is [-...]

F (x) = 4 times of SiNx + 4 times of cosx + 3 times of 2sinx - 2sinxcosx - 3 / 4 to find the minimum positive period

F (x) = (SiNx) ^ 4 + (cosx) ^ 4 + 2 (SiNx) ^ 3cosx-2sinxcosx-3 / 4 = 1-2 (sinxcosx) ^ 2 + (SiNx) ^ 2sin2x-sin2x-3 / 4 = - 1 / 2 (sin2x) ^ 2 + (1 - (COX) ^ 2) sin2x-sin2x + 1 / 4 = - 1 / 2 (1-cos4x) / 2-sin2x (1 + cos2x) / 2 + 1 / 4 = 1 / 4 (cos4x-sin4x) - 1 / 2 = I wrote

The minimum value of function f (x) = sin2x + 2sinx on [π, 2 π] is?

f'(x)=sin2x+2sinx=2cos2x+2cosx=4(cosx)^2-2+2cosx=4(cosx-1/2)(cosx+1)
When x0
Therefore, the minimum value of F (x) is f (5 π / 3) = sin10 π / 3 + 2sin5 π / 3 = - √ 3 / 2 - √ 3 = - 3 √ 3 / 2

It is known that f (x) = a · B − 1, where vectors a = (sin2x, 2cosx), B = (3, cosx), (x ∈ R). (1) find the minimum positive period and minimum value of F (x); (2) in △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If f (A4) = 3, a = 213, B = 8, find the value of side length C

(1) f (x) (1) f (x) is (1) f (x) = a (1) f (x) as (1) f (x) in (1) f (x) = a (1) f (x) in (1) f (x) as (1) f (x) is (1) f (x) in (1) f (x) as (1) f (x) in (1) f (x) is (1) B (1) f (x) = (2x, 2cosx, 2cosx) ((3, cosx) - 1 = 3s2x2x + 2cos2x (2, 2c2x, 2C (2) f (2x (2x, 2C (2) f (x (x) in (1) f (1) f (1) f (x (x (x) f (x (x) f (x) (1) f (x (x) f (x) (1) as (x (x) (1) f (x (x (x (x) in (x) (1) 2 or C = 6

Let f (x) =, where vector = (2cosx, 1), = (cosx, sin2x), X ∈ R
jessicahxf0131
1st floor
Let f (x) =, where vector = (2cosx, 1), = (cosx, sin2x), X ∈ R
(1) If f (x) = 0 and X ∈ (- π 2,0), find Tan 2x;
(2) Let a, B and C on the three sides of △ ABC form an equal ratio sequence in turn, and try to find the value range of F (b)

Is f (x) = AB-1? 1. F (x) = 2cos & # 178; X + sin2x-1 = cos2x + sin2x = √ 2Sin (2x + π / 4) = 0, so sin (2x + π / 4) = √ 2 / 2, X ∈ (- π / 2,0) so x = 0, so tan2x = tan0 = 02. B & # 178; = AC, f (b) = √ 2Sin (2B + π / 4) because CoSb = (A & # 178; + C & # 178; - B & # 178;) / (?)

Let f (x) = a * B, where vector a = (2cosx, 1), vector b = (cosx, (√ 3) sin2x), X ∈ R. (1) if f (x) = 0 and X ∈ (- π / 2,0), find tan2x; (2) let a, B, C on three sides of △ ABC be in equal proportion sequence, and try to find the value range of F (b)

f(x)=2cos^2 x+√3sin2x=1+cos2x+√3sin2x=2sin(2x+π/6)+1
(1) If f (x) = 0, sin (2x + π / 6) = - 1 / 2, X ∈ (- π / 2,0), x = - π / 6, 2x = - π / 3, tan2x = - √ 3
(2) B ^ 2 = AC, CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac ≥ 1 / 20