Find the period, monotone interval, maximum and minimum value of function y = sin (3x + π / 3), and write the set of independent variable x when reaching the maximum value and minimum value respectively

Find the period, monotone interval, maximum and minimum value of function y = sin (3x + π / 3), and write the set of independent variable x when reaching the maximum value and minimum value respectively

The period of the function y = sin (3x + π / 3) t = 2 π / 3
2kπ-π/2

y＝(sin x)＋cos X ∈ [(- pi / 2), PI / 2], the value range?

Take a radical 2, y = root 2Sin (x + π / 4), [- radical 2 / 2, radical 2 / 2]

The range of y = sin α - 3 / 2-cos α (α belongs to R) is solved by higher trigonometric function

2-cos α > 0, no matter what value α is, the fraction is always meaningful, and the domain of function definition is r
Organize, get
2y-ycosα=sinα-3
sinα+ycosα=2y+3
(1 + y? 2) sin (α + β) = 2Y + 3, where Tan β = y
sin(α+β)=(2y+3)/√(1+y²)
-1≤sin(α+β)≤1
-1≤(2y+3)/√(1+y²)≤1
(2y+3)²/(1+y²)≤1
Organize, get
3y²+12y≤-8
(y+2)²≤4/3
-2-2√3/3≤y≤2√3/3 -2
The value range of the function is [- 2 - 2 √ 3 / 3,2 √ 3 / 3 - 2]

2cos 30°-2sin 60°·cos 45°=? I hope you can answer in detail···

2cos 30°-2sin 60°cos 45°
=2*(√3/2)-2*(√3/2)*(√2/2)
=√3-√6/2

Given TaNx = 2, what is sin ^ 2x + (1 / 3) cos2x equal?

sinx^2+(1/3)(cosx^2-sinx^2)/sinx^2+cosx^2
=tanx^2+(1/3)(1-tanx^2)/tanx^2+1
It can be substituted

Given TaNx = 1 / 4, the value of cos2x + sin ^ 2x is

It is still a homogeneous evaluation
cos2x+sin^2x
=(cos²x-sin²x+sin²x)/1
=cos²x/(sin²x+cos²x)
=1/(tan²x+1)
=16/17

Given TaNx + 1 / TaNx = 2 / 5, X belongs to (π / 4, π / 2), calculate the values of cos2x and sin (2x + π / 4)

Because TaNx is greater than 1 on (π / 4, π / 2)
∵ TaNx is monotonically increasing on (π / 4, π / 2). If t = TaNx, then t ∈ (1, + ∞)
∵ T + 1 / T = 5 / 2 ᙽ T ^ 2-5 / 2T + 1 = 0 gives t = 2 or T = 1 / 2 (less than 1, round off)
∴tanx=2
On (π / 4, π / 2), cosx and SiNx are positive
From SiNx ^ 2 + cosx ^ 2 = 1 and SiNx / cosx = 2, SiNx = 2 √ 5 / 5, cosx = √ 5 / 5
∴cos2x=2cosx^2-1=-3/5,
√2sin(2x+π /4)=sin2x+cos2x=2sinxcosx+cos2x=1/5
∴sin(2x+π /4)=√2/10
The calculation process may be wrong, but the method should be like this

tanx=-2,π/2

sin(x+π/6) (sin2x+sin^2x)/(1-cos2x)
=sin(x+π/6)(2sinxcosx+sin^2x)/2sin^2x
=[1/2 +cosx/sinx]sin(x+π/6)
=[1/2+1/tanx]sin(x+π/6)
=0
I'm glad to solve the problem for you

The maximum and minimum values of the function y = cos2x cos ^ 2x-4sinx are also the values of X when the maximum and minimum values are taken

The solution is y = cos2x cos ^ 2x-4sinx
=1-2sin^2x-（1-sin^2x）-4sinx
=-sin^2x-4sinx
=-（sinx+2）^2+4
This function is a minus function in SiNx of [- 1,1]
So when SiNx = - 1, y has the maximum value y = - (- 1) ^ 2-4 (- 1) = 3
In this case, x = 2K π - π / 2, K belongs to Z
When SiNx = 1, y has a minimum value y = - (1) ^ 2-4 (1) = - 5
When x = 2K π + π / 2, K belongs to Z

In order to know the function f (x) = cos2x - (COS-1) cosx. (1) to find the minimum value of the function, (2) if x belongs to [2,3,2 of the permutation], compare f (x If we want to know the size of the permutation function (2) cos1 - (2), we can find the minimum value of (2) Cos2

F (x) = the square of cos2x cosx + cosx = 2cosx-1-cosx + cosx = cosx-1 + cosx
Let cosx = t
Original formula = square of T + T-1
Therefore, when t = - 1 / 2, the minimum value of F (x) is - 5 / 4
First question ~ second question, wait a moment~