The maximum value of function y = x + 2cosx radical 3 on [0, Pai / 2] If there is a space, it is "plus" seek

The maximum value of function y = x + 2cosx radical 3 on [0, Pai / 2] If there is a space, it is "plus" seek

First derivative: y '= 1-2sinx,
∵ x on [0, π / 2]
Let y '= 0,
X = π / 6
Then this point is a stationary point on [0, π / 2],
So when x = π / 6, there is a maximum and it is a maximum,
ymax=π/6

What is the maximum value of the function y = 3x + 2cosx on the interval [0,2 / Pie]?

Y '= 3-2sinx > 0, so y = 3x + 2cosx is an increasing function,
Thus, when x = π / 2, y has a maximum value of 0
3π/2+2cosπ/2=3π/2

The maximum value of the function FX = x + 2cosx on the interval [0, X / 2] is - - [0,2 π] is - [- π / 2,0 π]——

The first question is [0, π / 2]
F (x) = x + 2cos (x)
f'(x) = 1 - 2sin(x).
The first question is: on [0, π / 2], when x is less than 30 degrees, the derivative is positive and the function increases; when x is greater than 30 degrees, the derivative is negative and the function decreases;
Second, on [0,2 π], there are two zeros of derivative, one is obtained when x = π / 6, and it is a local maximum; the other is taken by x = 5 π / 6, and the monotonicity of the function is as follows:
[0, π / 6], the derivative is positive and the function increases monotonically;
(π / 6,5 π / 6], the derivative is negative and the function decreases monotonically;
(5 π / 6,2 π], the derivative is positive and the function increases monotonically
Therefore, the maximum value should be compared with two points, one is at π / 6 and the other is at 2 π. Since f (2 π) = 2 π + 2 > F (π / 6), the answer to the second question is: 2 π + 2;
Third, I don't know which interval you asked
However, what is certain is that the function is positive on [- π / 2, π / 6), and it is monotonically increasing. If π / 6 can be taken, then the maximum value is f (π / 6). In short, if the interval of the question includes 2 π, then the maximum value is the same as the second question. If not, you can judge it by derivative yourself. It is very easy

The function y = x + 2cosx in [0, π 2] The value of X is () A. 0 B. π Six C. π Three D. π Two

y′=1-2sinx=0  x∈[0，π
2]
The solution is: x = π
Six
When x ∈ (0, π)
6) When y ′ > 0,  function is at (0, π)
6) On monotone increasing
When x ∈ (π)
6，π
2) When y ′ < 0,  function is at (0, π)
6) Monotonically decreasing,
The function y = x + 2cosx is in [0, π
2] X = π when the maximum value is obtained on
Six
Therefore, B

When y = 2cos2x-3x takes the maximum value, TaNx =?

The subject is wrong
Because this function has no maximum,
For example, X tends to be negative infinity and Y tends to be positive infinity

Given that α is an obtuse angle Tan (α + π / 4) = - 1 / 7, find the value of (1) Tan α and (2) find the value of Cos2 α + 1 / √ 2 cos (α - π / 4) - Sin 2 α (2) Find the value of Cos2 α + 1 / √ 2 cos (α - π / 4) - sin2 α. The tick is the root sign.

tan（α+π/4）=(tanα+tanπ/4)/(1-tanα*tanπ/4)=(tanα+1)/(1-tanα)=-1/7
tanα= -4/3
cos2α=[1-tan^2α)]/[1+tan^2α] =(1-(-4/3)^2)/(1+(-4/3)^2)=-7/25
sin2α=2tanα/(1+tan^2α)=2*(-4/3)/(1+(-4/3)^2)=-24/25
cosα=-3/5 sinα=4/5
cos2α+1/√2 cos(α-π/4）-sin2α= cos2α+cosπ/4 cos(α-π/4）-sin2α
=cos2α+cosπ/4 cos(α-π/4）-sin2α=cos2α+(cosα+ cos(π/2-α))/2-sin2α
=cos2α+(cosα+ sinα)/2-sin2α=-7/25+(-3/5+4/5)/2-24/25=-57/50

Trigonometric function problem: the maximum value of F (x) = cos ^ 2 (x) + 3sinx * cosx is, please write the detailed process

f(x)=(1+cos2x)/2+3/2*sin2x
=3/2*sin2x+1/2*cos2x+1/2
From the auxiliary angle formula
=√[(3/2)²+(1/2)²]sin(x+z)+1/2
=√10/2*sin(x+z)+1/2
Where Tanz = (1 / 2) / (3 / 2) = 1 / 3
Only the maximum value = √ 10 / 2 + 1 / 2

On the problem f (cosx) = x / 2, find f (COS (4 π / 3)) 1.f(cosx)=x/2 (0

1. Cos (4 π / 3) = cos (2 π - 2 π / 3) = cos (2 π / 3), so f (COS (4 π / 3)) = f (COS (2 π / 3)) = (2 π / 3) / 2 = π / 32, f (SiNx) = 3-cos2x = 3 - (1-2 (SiNx) ^ 2) = 2 + 2 (SiNx) ^ 2 F (cosx) = f (radical 1 - (SiNx) ^ 2

f（x）=2sin²x-λcosx+5 The maximum value of F (x) = 2Sin? X - λ cosx + 5 in [0, π / 2] is 5?

If f (x) = 2Sin? X - λ cosx + 5 = 2-2cos? X - λ cosx + 5, let cosx = t, f (T) = - 2T? - λ T + 7, the value range of T is [0,1], and then the solution is based on the idea of finding the maximum value of quadratic function

If cos (5 π / 12 + α) = 1 / 3, - π < α < - π / 2, then cos (π / 12 - α) =? If cos (5 π / 12 + α) = 1 / 3, - π < α < - π / 2, then cos (π / 12 - α) =? (- 2 root sign 2 / 3) cos（π/12-α）=sin(5π/12+α ) Export metropolis is to ask the final answer why there are signs and how to look at the scope

cos（π/12-α）
= sin(π/2-(π/12-α) )
=sin(5π/12+α )
Because - π < α < π / 2,
So - 7 π / 12