The solution of the equation SiNx / 2 = 1 / 3 in [π, 2 π] x =?

The solution of the equation SiNx / 2 = 1 / 3 in [π, 2 π] x =?

∵x∈[π,2π]
∴x/2∈[π/2,π]
∵sinx/2=1/3
∴x/2=π-arcsin(1/3)
∴x=2π-2arcsin(1/3)

Tangent equation of y = SiNx at x = 1 / 2
F (x) = alnx - (1 + a) x + 1 / 2, a &; R
solve

y'=cosx
When x = 1 / 2,
y=sin(1/2),y'=cos(1/2)
Then the tangent at this point is y = cos (1 / 2) * (x-1 / 2) + sin (1 / 2)

Let f (x) satisfy f (- SiNx) + 3f (SiNx) = 4sinxcosx (| x | ≤ half π) to find the expression of F (x)

If f (x) satisfies f (- SiNx) + 3f (SiNx) = 4sinxcosx (| x | ≤ 2 / 2 π) (1) replace x with - x, then f [- sin (- x)] + 3F [sin (- x)] = 4sin (- x) cos (- x), that is, f (SiNx) + 3f (- SiNx) = - 4sinxcosx (| x | ≤ 2 / 2 π) (2) (1) * 3 - (2)

It is proved that the equation 1 + X + X & sup2 / 2 + X & sup3 / 6 = 0 has only one real root
Using Rolle's mean value theorem to prove

Let f (x) = 1 + X + X & sup2 / 2 + X & sup3 / 6
Because f (0) = 1 > 0, f (- 2) = - 1 / 30
F '(ξ) = 0 and contradiction
So f (x) has only one real root

How to do the equation of 0.8x-2 = 0.6x + 6

Transference
0.8x-0.6x=6+2
0.2x=8
x=8÷0.2
x=40

It is proved by Rolle's theorem that the equation SiNx + xcosx = 0 must have real roots in (0, π)

F (x) = xsinx, f (0) = f (PI) = 0. According to Rolle's mean value theorem, there exists C such that f '(c) = 0, f' (c) = sinc + ccosc = 0,

Using Rolle's theorem to prove that x Λ 5 + X-1 = 0 has only one positive root

Using the intermediate value theorem, it is proved that the equation x & sup3; + X-1 = 0 has and only has one real root

First, prove the existence of real roots, Let f (x) = x ^ 3 + X-1, then f (0) = - 1, f (1) = 1, according to the intermediate value theorem, there is a real root t between (0,1), such that f (T) = 0. Second, prove the uniqueness, suppose there are two unequal real roots, let the two real roots be m and n (M is not equal to n) (M and N are between (0,1), then there is

It is proved that the equation X-2 ^ x = 1 has at least one positive root less than 1

It is proved that the equation X-2 ^ x = 1 has at least one positive root less than 1
Prove: ∵ equation X-2 ^ x = 1
Let f (x) = X-2 ^ X-1
Let f '(x) = 1-2 ^ xln2 = 0 = = > 2 ^ x = 1 / LN2 = = > x = ln (1 / LN2) / LN2 = - ln (LN2) / LN2
f’’(x)=-2^x(ln2)^2

It is proved that one of the two roots of the equation (x-a) (x-a-b) = 1 is greater than a and the other is less than a

Proof: let x-a = y
∵(x-a)(x-a-b)=1
∴y(y-b)=1
y^2-by-1=0
Let the two real roots of the equation y ^ 2-by-1 = 0 be Y1 and Y2 respectively
∵y1*y2=c/a=-1＜0
{Y1 and Y2 are different signs
∵y=x-a
∴y1=x1-a＜0,y2=x2-a＞0
∴x1＜a,x2＞a
Or Y1 = x1-a > 0, y2 = x2-a < 0
∴x1＞a,x2＜a
The two equations (x-a) (x-a-b) = 1 about X, one of which is greater than a and the other is less than a