explain
y =[2(cosx- 1)+ 1]/(cosx- 1)= 2+ 1/(cosx- 1)
The domain is: cosx- 1 not =0, that is, cosx is not = 1.
That is, x is not equal to =2kπ.
That is, the domain is {x|x belongs to r, and x does not =2kπ}.
Because-1
That is,1(cosx-1) < =-1/2.
Then: 2+ 1/(cosx- 1)
That is, the value range of y is (-infinity, 3/2).
2 It is known that the solution set of the equation cosx=(2a- 1)/5 about X is an empty set, so find the value range of A. ..
The answer is because-1
(2a- 1)/5 & lt； -1 or (2a-1)/5 >; 1.
Namely: 2a-1; five
get:a & lt； -2 or a> three
3 Given that the maximum value of the function f(x)=asinx+bcosx(a, b is not equal to 0) is 2, and f(π/6)= radical number 3, find the value of f(π/3).
f(x)=asinx+bcosx
= radical sign (a 2+b 2) sin (x+m)
If the maximum value is 2, then the root sign (A 2+B 2) = 2.
That is f(x)=2sin(x+m).
F(π/6)= root number 3.
Namely: √3=2sin(π/6+m)
π/6+m=π/3
m=π/6
So: f(π/3)=2sin(π/3+π/6)=2.
The equation about x is known. x？ -(tanα+cotα)x+ 1=0 has a root of 2+ a root number of 3. Find the value of sin2α.
explain
Set x? -(tanα+cotα)x+ 1=0 followed by x 1=2+√3 and x2 respectively.
According to Vieta's theorem, x 1x2= 1, x2 =1/x1=1(2+√ 3) = 2-√ 3.
The decomposable factor of the original equation is [x-(2-√3)][x-(2+√3)]=0.
Expansive x? -4x+ 1=0
Compared with the coefficient in the original equation, tanα+cotα=4.
tanα+cotα= tanα+ 1/tanα=( 1+tan？ α)/tanα= seconds? α/tanα= 1/sinαcosα= 2/sin 2α= 4
sin2α= 1/2