If t = SiNx + cosx, and sin ^ 3x + cos ^ 3x

If t = SiNx + cosx, and sin ^ 3x + cos ^ 3x

It's not hard to do this:
∵ SiNx + cosx = t, (SiNx + cosx) & sup2; = T & sup2;, so
sinx*cosx=(t²-1)/2；
The original equation sin ^ 3x + cos ^ 3x = (SiNx + cosx) (SiNx & sup2; - SiNx * cosx + cosx & sup2;)
=t*(1-(t²-1)/2)
So t * (1 - (T & sup2; - 1) / 2) > 0 solves the inequality
(- root 3,0) ∪ (root 3, + infinity) (1)
And T = SiNx + cosx = root 2 (sin (x + π / 4)), so
-Root 2
If the root cos ^ 2x = cosx, then the value range of X is
Root cos ^ 2x = cosx ≥ 0
-π/2+2kπ≤x≤π/2+2kπ (k∈Z)
If sin ^ 2x > cos ^ 2x, then the value range of X is?
As shown in the figure, ox is the starting edge of the angle, and when the angle t satisfies Sint > cost, its final edge should be in the area where the ray OA rotates counterclockwise to ob, that is, 2K π + π / 4
Range of logarithm function
The range of y = Log1 / 2 (- x ^ + 2x + 3) is
(-x^2+2x+3)>0
x^2-2x-3
(-x^+2x+3)>0
x^-2x-33 ,(-x^+2x+3)-->0, y-->+∞
x=1,y=-2
Range [- 2, + ∞)
dsdsd
First (- x ^ 2 + 2x + 3) use junior high school matching method or other methods
The solution (- x ^ 2 + 2x + 3) is less than or equal to 4
And (- x ^ 2 + 2x + 3) is greater than 0
Then Y > Log1 / 2 (4) = - 2
So Y > - 2
If f (x) = LG (2x1 + X + a) (a ∈ R) is an odd function, then a=______ .
This is one of the odd functions that you want to get (x (x) = LG (2x1 + X + 1 + X + a) (a ∈ R) is one of the odd functions that you want to want to get (2 + 2 + a) to get (2x1 + X + 1 (2x1 + X + 1 + X + a) (a \ (x) = LG (2x1 + X + X + a) is the odd function that you want to get (2 + 2 + a) this is one of the most common and the odd function that is the odd function that you want \ \\\\\\\\\\\\\\\\\\\\\\\\\= - 1, so the answer is: - 1
The range of logarithm function
Logarithm function loga (9-6a). The range of a is 1
a> 1, so y = logax is an increasing function,
Because 1
The function f (x) defined on the interval (- 1,1) satisfies 2F (x) - f (- x) = LG (x + 1), and the analytic expression of F (x) is obtained
Why can 2F (- x) - f (x) = LG (- x + 1) be obtained from 2F (x) - f (- x) = LG (- x + 1)?
Because 2F (x) - f (- x) = LG (x + 1), --- (1) x is defined on (- 1,1), it is also true to replace x with - x, that is, 2f (- x) - f (x) = LG (1-x) -- (2) (2) left and right sides of formula (1) multiply by 2, then f (- x) can be eliminated, and 3f (x) = 2 * LG (1 + x) + LG (1-x)
Logarithmic function domain solution (online, etc.)
(1) Definition field and value field of loga (x ^ 2)
(2) Log a (9-x ^ 2) definition field and value field
（1）{x|x≠0},R
（2）(-3,3),
Range: because 9-x ^ 2 ∈ (0,9],
When a > 1, the range: (- ∞, loga (9)];
When 0
Discussion on the value of a!!
x ≠ 0
-3
If the function f (x) defined on the interval satisfies 2F (x) - f (- x) = LG (x + 1), then the analytic expression is
2f(-x)-f(x)=lg(-x+1)
4f(x)-2f(-x)=2lg(x+1)
Two formula addition
3f(x)=lg(x+1)^2*(1-x)
f(x)=[lg(x+1)^2*(1-x)] /3
How to calculate the logarithm function log in senior one's mathematics with computer?
Use matlab to calculate! Ln stands for log, eg: log2 can be directly input into LN2!