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

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] The maximum value of π is obtained
Six
Therefore, B

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

Let SiNx = t
y=2(1-2t²)-3t
=-4t²-3t+2
Image opening down,
That is, t = - 3 / 8, y has a maximum,
At this point, SiNx = - 3 / 8
cosx=±√55/8
tanx=±3/√55=±3√55/55

Given a vector a = (SiNx, SiNx + cosx) B = (2cosx, cosx SiNx), Let f (x) = a * B (1) When x ∈ [0, π / 2], find the value range of function f (x); (2) Find the value of sin (2,4) = (2,4) / (2,4) = (2,4) = (2,4) / (2) = (2) / (2) = (2) / (2) = (2) / (2) = (2) / (2) = (2) / (2) = (2) = (2) = (

If x ∈ [0, π / 2], then 2x + π / 4 ∈ [π / 4, π / 4 + π] when x ∈ [0, π / 2], 2x + π / 4 ∈ [π / 4, π / 4 + π] when 2x + π / 4 = π / 2 = π / 4, π / 4 + π] when 2x + π / 4 = π / 2, f (x) takes the maximum value, when 2x + π / 4 = - π / 2, f (x) takes the minimum value, so the value of F (x) is [- √, so the value domain of F (x) is [- / [- / / / - / / - / - / - / - - / (x) is the value domain of F (x) is 2

Let s (x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, X, x, x, x, x, x, x, x, x, x, x, x, x, X (1) Find the monotone increasing interval of the minimum positive period sum of function f (x) on degree [0180] (2) When x belongs to [0,30 degrees], - 4

F (x) = 2cos ^ 2 + radical 3sin2x + M
=Cos2x + Radix 3sin2x + M-1
=2sin(2x+π/3)+M-1
So t = 2 π / 2 = Pi
2X + Pai / 3 belongs to [2K sect / 2,2k + Pai / 2]
Because x belongs to [0180]
So the increasing interval {π / 3, π / 2] ∪ [3 π / 2,7 π / 3]
2) 2X + π / 3 belongs to Pai / 3,2 π / 3
So sin (2x + π / 3) belongs to [radical 3 / 2,1]
Therefore, f (x) belongs to [M + radical 3-1, M + 1]
So m + 1-4
So m belongs to [- radical 3-3,3]

Given the vector a = (SiNx, cosx), B = (SiNx, - root sign 3sinx), (1) find the function f (x) = the monotone increasing interval of vector a * vector B, (2) can the image of F (x) be transformed from the image of y = SiNx? Please write the transformation process

Given the vector a = (SiNx, cosx), B = (SiNx, - root sign 3sinx), (1) find the function f (x) = the monotone increasing interval of vector a * vector B, (2) can the image of F (x) be obtained from the image transformation of y = SiNx? Please write the transformation process (1) analysis: ∵ vector a = (SiNx, cosx), B = (SiNx, - 3sinx)

Let a = (SiNx, cosx), B = (SiNx, √ 3sinx), X belongs to R, and the function f (x) = a (a + 2b) Let a = (SiNx, cosx), B = (SiNx, Radix 3sinx), X belong to R, and the function f (x) = a (a + 2b) 1. Find the minimum positive period T of F (x) 2. Given that a, B, C are the opposite sides of the inner angles a, B, C of △ ABC, where a is an acute angle, a = √ 3, C = 2, and f (a) is exactly the maximum value of F (x) on [0, the school of half]. Find the values of angle A and edge B

(3sinx, cosx + 2 √ 3sinx) is an important part of the system, and it can be used to analyze (SiNx, cosx) (3sinx, cosx + 2 √ 3sinx) = 3sinx + cos (x + 2 + 2 √ 3sinx) x + 2 √ 3sinxcos x = 1 + 2 SiNx + 2 + 2 √ 3 SiNx cosx = 1 + 1-cos2x + √ 3sin22x = 2 + 2Sin (2x - π / 6) f (x) f (x) the minimum positive cycle T = = minimumpositive cycle of the minimum positive cycle T = (x) (2x - π / 6) f (x) f (x) f (x) has the minimum positive cycle T = (t = (x) the minimum positive cycle of the minimum positive cycle T = and

Radix 1-2 sin10 degrees cos10 degrees divided by sin10 degrees - Radix 1-sin10 degrees squared

√1－2sin10cos10
————————
sin10－√1－sin②10
√sin②10+cos②10－2sin10cos10
＝——————————————
sin10－√sin②10+cos②10－sin②10
√（sin10－cos10）②
= ——————————
sin10－cos10
sin10－cos10
= ——————
sin10－cos10
=1 Baidu map

Reduce the square of (sin10 - cos10) under root sign

The square of (sin10 - cos10) under the radical
=|sin10°-cos10°|
=cos10°-sin10°

Simplification: 1 + cos 100 degree * sin 170 degree divided by cos 370 degree + 1-sin square 170 degree under root sign

1 + cos 100 degree * sin 170 degree divided by cos370 degree + 1-sin square 170 degree under root sign = 1 + (- Sin 10 degree) * sin 10 degree divided by cos370 degree + 1-sin square 10 degree under root sign = 1-sin square 10 degree under root sign divided by cos10 degree + cos square 10 degree under root sign = cos square 10 degree divided by cos10 degree

Reduction process of radical 2 / 2cos (2x + π / 4) + sin ^ 2 x Radical 2 / 2cos (2x + π / 4) + sin ^ 2 x = 1 / 2cos2x-1 / 2sin2x + 1 / 2 (1-cos2x) = 1 / 2-1 / 2sin2x, thank you

cos(2x+π/4）=cosπ/4cos2x-sinπ/4sin2x
Cos π / 4 = sin π / 4 = root of 2
cos2x=cosx^2-sinx^2=1-2sinx^2
sinx^2=(1-cos2x)/2