How does 1 / 2cos2x-2 / Radix 3sin2x be combined and deformed

How does 1 / 2cos2x-2 / Radix 3sin2x be combined and deformed

1 / 2 * cos2x - root of 2 / 3sin2x
=cos2x*cos(π/3) - sin2x*sin(π/3)
=cos(2x+ π/3)

Using trigonometric function line to find the definition domain of y = LG (2cosx radical 3)

2cosx-√3>0
cosx>(√3)/2>0
Make a straight line perpendicular to the Y axis through a point on the unit circle
cosα=x/r
In this hypothesis, r = 1
cosα=(√3)/2
α = 2K π + π / 6 or α = 2K π - π / 6
x> (√ 3) / 2
2kπ-π/6≤x≤2kπ+π/6

The definition domain of trigonometric function plus sinx-2

If you are wrong, it should be the root of sin (X-2)
2kπ+2〈x〈2kπ+2+π

Function y = lgsin2x+ The definition domain of 9-x2 is___ .

∵ function y = lgsin2x+
9-x2，
ν
sin2x＞0
9-x2≥0 ，
The solution
kπ＜x＜(k+1
2)π
-Where (x ≤ 3);
∴-3≤x＜-π
2, or 0 ＜ x ＜ π
2；
The definition domain of the function is [- 3, - π
2）∪（0，π
2）；
So the answer is: [- 3, - π
2）∪（0，π
2）．

What is the definition domain of the square of 9-x under y = radical?

Greater than or equal to 0 in root sign
Yes
9-x^2 >=0
>
-3

The square of X under y = root sign - 4x + 9 to find the function definition domain

Y = radical (x ^ 2-4x + 9) = radical [(X-2) ^ 2 + 5]
When x is any real number, (X-2) ^ 2 + 5 is greater than zero

What is the definition domain of the function y = the square of 1-x + the square of X-1?

If the algebraic expression under the radical is greater than or equal to 0, X (x-1) is greater than or equal to 0, and the answer is that x is less than or equal to 0 or X is greater than or equal to 1

Function y = 1 The domain of 6 − x − X2 is______ .

To make the function y = 1
The analytic formula of 6 − x − X2 is meaningful
The independent variable x must satisfy:
6-x-x2＞0
That is, X2 + X-6 < 0
The solution is - 3 < x < 2
So the function y = 1
The domain of 6 − x − X2 is (- 3, 2)
So the answer is: (- 3, 2)

50 math problems with radical Find 50 math problems with root sign

1. 2√12*√3
2. √18/√2
3. （√3-2）*（√3+2）
Something like this

A mixed mathematical problem with radical The second power of (Part A under the root B + part a under the root b)

The original formula = B / A + 2 + A / b = (b ^ 2 + A ^ 2) / AB + 2 = (a ^ 2 + 2Ab + B ^ 2) / AB = (a + b) ^ 2 / AB