What is the proof of mathematical HL

What is the proof of mathematical HL

If the hypotenuse of two right triangles is equal to that of one right triangle, the two triangles are congruent
Is that it?

The solution set of the equation sin ^ 2x minus 2sinx = 0 is? And the answer is {x | x = KX,

sin^2x-2sinx=0
sinx(sinx-2)=0
SiNx = 0 or SiNx = 2 (rounding off)
∴sinx=0
X = KX, K belongs to Z

Let the equation sin (2x + π 6) = K + 12 have two different roots α, β in [0, π 2]. Find the value of α + β and the range of K

∵ x ∈ [0, π 2], ∵ (2x + π 6) ∈ [π 6, 7 π 6]. ∵ the equation sin (2x + π 6) = K + 12 of X has two different roots α, β, ∵ 12 = sin π 6 ≤ K + 12 < 1 in [0, π 2]. The solution is 0 ≤ K < 1, ∵ α + β = 2 × π 2 = π

If the equation cos2x + sinx-a = 0 of X has a real solution, then the minimum value of real number a is______ ．

∵ cos2x + sinx-a = 0 ∵ - sin2x + SiNx + 1-A = 0 is equivalent to: - Y2 + y + 1-A = 0 & nbsp; & nbsp; & nbsp; ∵ a = - (Y-12) 2 + 54 ∵ y ∈ [- 1, 1] ∵ - (Y-12) 2 ∈ [- 94, 0], that is, the minimum value of a ∈ [- 1, 54] ∵ A is: - 1

If a is known to belong to [- 3,3], then the equation cos2x + 2sinx-a = 0 about X has a solution in the interval [- π / 6, π]

x∈[-π/6,π]
sinx∈[-1/2,1]
cos2x+2sinx-a=0
1-2sin²x+2sinx-a=0
2sin²x-2sinx+a-1=0
(sinx-1/2)²=(3-2a)/4
sinx∈[-1/2,1]
0≤(sinx-1/2)²≤1
0≤(3-2a)/4≤1
-1/2≤a≤3/2
|3/2-(-1/2)|/|3-(-3)|=1/3
The probability is 1 / 3

Given that the equation cos2x + 4sinx-a = 0 has a solution, then the value range of A

The original equation can be transformed into a = cos2x + 4sinx = 1-sin2x + 4sinx = - (sinx-2) 2 + 5 ∵ SiNx ∈ [- 1, 1] ∵ a ∈ [- 4, 4]

Given that the equation cos2x + 2sinx + 2a-3 = 0 has exactly two real roots in (0,2 π), the value range of a is obtained
What's the first step?

3/4

If the equation √ 3sin2x + cos2x = k about X has two different real roots α, β in the interval [0, π / 2], find the value range of real number k and α + β

3sin2x+cos2x=k
sin(2x+π/6)=k/2
It is easy to find that f (x) = sin (2x + π / 6) is an increasing function on [0, π / 6] and a decreasing function on [π / 6, π / 2]
If sin (2x + π / 6) = K / 2 has two different real roots α, β, because 0 is close to the opposite axis X = π / 6,
Then f (0)

It is known that the equation cosx-sinx-2sinx + 2A + 1 = 0 has a solution in the interval (0, π / 2). Find the value range of real number a

=0 take your time

If there is a real number x satisfying cosx sin ^ 2 = a, then the value range of real number a is___

Find the range of cosx sin ^ 2, (first change the unified trigonometric function, then use it between - 1,1) a is the range of the range