Let f (x) = root 2 of log 2 (x of root 2 sin 2), Find the definition domain, the minimum value, and the coordinates of the intersection point with the x-axis

Let f (x) = root 2 of log 2 (x of root 2 sin 2), Find the definition domain, the minimum value, and the coordinates of the intersection point with the x-axis

Definition domain: sin (x / 2) > 0, namely 2K π

Let 2 ^ x = 1 / 2 be known. Find the maximum and minimum values of (x / 2) * log root 2 of the function f (x) = log2

2^x≤256=2^8
x≤8
Log (2) x ≥ 1 / 2, X ≥ √ 2
So the range of X is [√ 2,8]
(x / 2) * log √ 2 of function f (x) = log2 (√ X / 2)
=Log2 (x / 2 * x / 4)
=Log2 (x ^ 2 / 8)
When x = √ 2, f (x) min = - 2
When x = 8, f (x) max = 3

How to deduce the formula sin ^ 2x = (1-cos2x) / 2 cos ^ 2x = (1 + cos2x) / 2?

It's not something new. It's just a variation of the formula for double angle
cos2x=1-2sin^2x=2cos^2x-1
You'll find out if you take a closer look

Given 4sin? X-cos? X-6sinx + 3cosx = 0, find the value of (cos2x cos? X) / 1-cot? X

4sin? X-cos? X-6sinx + 3cosx = 0, (2sinx) ^ 2 - cos ^ 2x - 3 (2sinx - cosx) = 0 (2sinx - cosx) * (2sinx - cosx - 3) = 02sinx - cosx = 0, or 2sinx - cosx - 3 = 0 (this does not hold) 2sinx = cosx, Cotx = cosx / SiNx = 2Sin ^ 2x +

It is known that 4sin Λ 2x-6sinx cos Λ 2x + 3cosx = 0

For what

It is known that 4sin? X-cos? X-6sinx + 3cosx = 0, Find the value of (cos2x cos? X) / (1-cot? X) Explain it clearly

(2sinx+cosx)(2sinx-cosx)-3(2sinx-cosx)=0
(2sinx-cosx)(2sinx+cosx-3)=0
Because of SiNx

y=lim (cos2x)^（1+cot^2x） y=lim(x→0) (cos2x)^（1+cot^2x） =lim(x→0) e^ln(1-2xin^2x)/xin^2x What formula of LN is used in this book,

The logarithm with base e is usually represented by y = INX (x as the independent variable and y as the logarithm of X with E as the base)

The maximum value of the function y = cos2x + 2sinx is______ .

∵ the function y = cos2x + 2sinx = - 2sin2x + 2sinx + 1 = - 2 (SiNx − 1)
2)2+3
2，
When SiNx = 1
When 2, the maximum value of function y is 3
2，
So the answer is: 3
2．

The value range of the function y = cos2x-2sinx is () A. [-3，1] B. [-3，3 2] C. [-1，1] D. [3，3 2]

∵ function y = cos2x-2sinx = 1-2sin2x-2sinx = 3
2-2 (sinx+1
2)2，
When SiNx = - 1
2, the maximum value obtained by the function is 3
2. When SiNx = 1, the minimum value of the function is - 3,
Therefore, the value range of the function is [- 3,3]
2]，
Therefore, B

The value range of the function y = cos2x cosx + 5 / 4 is The two of cos2x is square

The original formula = cos ^ 2x cosx + 1 / 4 + 1
=(cosx-1/2)^2+1
It's not hard to see that this is the form of a quadratic function
When cosx = 1 / 2, the minimum value of the function is 1
Because the cosx range is [- 1,1]
So compare the values of the functions at 1 and - 1, and take the larger one
f(1)=5/4,f(-1)=13/4
So the range of the function is [1,13 / 4]