Given 4sin square x-6sinx-cos square x+3cosx=0, X ∈ (0, π /2), find (1) cosx (2) (5sinx-3cosx) / (cosx+sinx)

Given 4sin square x-6sinx-cos square x+3cosx=0, X ∈ (0, π /2), find (1) cosx (2) (5sinx-3cosx) / (cosx+sinx)

4sin square x-6sinx cos square x + 3cosx = 0
(2sinx+cosx)(2sinx-cosx)-3(2sinx-cosx)=0
(2sinx-cosx)(2sinx+cosx-3)=0
2sinx cosx = 0 (1) 2sinx + cosx-3 = 0 (no solution)
sin²x+cos²x =1 (2)
sinx>0 cosx>0 (3)
∴sinx=√5/5 cosx=2√5/5
(5sinx-3cosx)/（cosx+sinx） =(√5-6√5/5)/(2√5/5+√5/5)=-1/3

The value range of the function y = cos2x-2sinx (- π / 4 ≤ x ≤ π / 4) is The point is (- π /4 ≤ x ≤π /4) ah

y=cos2x-2sinx=-2sinx^2-2sinx+1=-2(sinx+1/2)^2+3/2
-π/4≤x≤π/4→-√2/2≤sinx≤√2/2
When SiNx = - 1 / 2 (x = - π / 6), ymax = 3 / 2
When SiNx = √ 2 / 2 (x = π / 4), Ymin = - √ 2

The value range of the function y = cos2x-8cosx is______ .

Y = cos2x-8cosx = 2cos2x-8cosx-1 = 2 (cosx-2) 2-9, because cosx ∈ [- 1, 1], and when cosx < 2, y is a decreasing function, so when cosx = 1, the minimum value of Y is 2 × (1-2) 2-9 = - 7; when cosx = - 1, the maximum value of Y is 2 × (- 1-2) 2-9 = 9

10. The value range of function y = | cos2x | + | cosx | is () A.[12,2] B.[22,2] C.[22,98 ] D.[32,2] This is the tenth question of Science in Wuhan in February 2007 A [0.5,2] B [radical 2 / 2, 2] C [radical 2 / 2,9 / 8] d [radical 3 / 3,2]

It is required that the range of values should be simplified first, without looking at the absolute value
y=2(cosx)^2-1+cosx
Using the commutation law t = cosx (- 1 =)

The value range of function cos2x + cosx + 1

cos2x+cosx+1=2(cosx)^2+cosx=2(cosx+1/4)^2-1/8
The range of cosx is [- 1,1]
When cosx = - 1 / 4, the minimum value is - 1 / 8
When cosx = 1, the maximum value is 3
So the range is [[- 1 / 8,3]

When x → 0, LIM (1 + x ^ 2) ^ cot ^ 2x needs a detailed process I can't type all the mathematical symbols of the title. When x tends to 0, find the [cot square x] power of the whole [1 plus x squared]

lim(1+x²)^cot²x
=lim(1+x²)^(1/x²)(x²cot²x)
=lim e^(x²/tan²x)
=e

The value range of the function y = cos2x-cosx-1 is_____

y=2cos²x-1-cosx-1
=2(cos-1/4)²-17/8
-1

1. Given SiNx / 2 - cosx / 2 = - √ 5 / 2, find the value of SiNx and cos2x? 2 y = LG (1 + TaNx)?

1. SiNx / 2 - cosx / 2 = - √ 5 / 2,
(sinx/2)^2+(cosx/2)^2-2sinx/2*cosx/2=5/4
And (SiNx / 2) ^ 2 + (cosx / 2) ^ 2 = 1,2sinx / 2 * cosx / 2 = SiNx
The deformation of the above formula is 1 + SiNx = 5 / 4, so SiNx = 1 / 4
cos2x=1-2(sinx)^2=1-2*(1/4)^2=1-1/8=7/8
2. The function satisfies 1 + TaNx > 0, that is, TaNx > - 1 = Tan (K π - π / 4)
Therefore, X belongs to the interval (K π - π / 4, K π + π / 2), which is the definition domain of y = LG (1 + TaNx)

Know TaNx = 2, how to find tan2x =?

Tan2x = 2tanx / (the square of 1-tanx) = 2 * 2 / (1-4) = - 4 / 3

The minimum positive period of y = 1 / 2sin2x + root 3cos ^ 2x - radical 3 / 2 is

=1 / 2sin2x + root 3 / 2 (2cos ^ 2x-1)
=1 / 2sin2x + root 3 / 2cos2x
=sin2xcos60+cos2xsin60
=sin(2x+60)
Minimum positive period T = 2 Π / 2 = Π