arcsin(3x-4x^3) -3arcsinx = 0.

We use the fact that f(f^-1(x) ) = x .

We know sin2A = 2sinAcosA.

Cos2A = 1-2sin^2A

Sin 3A = sin(2A+A) = sin2A*cosA +cos2A*sinA

Sin3A = 2sinA*cos^2A + (1-2sin^2A)sinA.

Simplifying we get:

sin 3A = 3sinA - 4sin^3A is a trigonometric identity.

We use the above properties in establishing the identity.

Substitute x = siny in the given equation:

arcsin (3siny -sin^3y) = 3 arcsin (siny) = 0 ,

arcsin (sin3y) - 3 arcsin (siny) = 0, as 3siny -4sin^3y = sin 3y.

3y - 3y = 0, becuse of the property of inverse function.

Therefore , the given identity stands proved.

We'll apply Lagrange's consequence.

We'll note the expression from the left side as the function f(x).

f(x) = arcsin(3x-4x^3)-3arcsinx

f(x) = 0

We'll calculate the first derivative of the function, both sides:

[arcsin(3x-4x^3)-3arcsinx]' = (0)'

(arcsin x)' = 1/sqrt(1-x^2)

[arcsin(3x-4x^3)]' = (3x-4x^3)'/sqrt[1 - (3x-4x^3)^2]

[arcsin(3x-4x^3)]' = (3 - 12x^2)/sqrt[1 - (3x-4x^3)^2] (1)

(3arcsinx)' = 3/sqrt(1-x^2) (2)

[arcsin(3x-4x^3)-3arcsinx]' = (1) - (2)

3(1 - 4x^2)/sqrt[1 - (3x-4x^3)^2] - 3/sqrt(1-x^2) =

= 3(1 - 4x^2)sqrt(1-x^2)-3sqrt[1 - (3x-4x^3)^2]/sqrt[1 - (3x-4x^3)^2]*sqrt(1-x^2)

If (1) - (2) = 0, then the identity is true!

We have to prove that arcsin(3x-4x^3)-3arcsinx=0

Here we first take a function

f(x) = arcsin(3x-4x^3) - 3arcsinx

and as it is given that arcsin(3x-4x^3) - 3arcsinx=0

f(x) = 0.

Now let us differentiate the function.

=> f'(x) = [arcsin(3x-4x^3)]' - [3arcsinx]'

We know that (arcsin x)' = 1/sqrt(1-x^2)

Also, [arcsin(3x-4x^3)]' = (3x-4x^3)' / sqrt[1 - (3x-4x^3)^2]

=> (3 - 12x^2) / sqrt[1 - (3x-4x^3)^2]

Therefore f'(x) = [arcsin(3x-4x^3)]' - [3arcsinx]'

=> 3(1 - 4x^2) / sqrt[1 - (3x-4x^3)^2] - 3/sqrt(1-x^2)

=> 3(1 - 4x^2) sqrt (1-x^2) - 3sqrt [1 - (3x-4x^3)^2] / sqrt[1 - (3x-4x^3)^2] *sqrt (1-x^2)

Now as

f'(x) = 3(1 - 4x^2) sqrt (1-x^2) - 3sqrt[1 - (3x-4x^3)^2] / sqrt[1 - (3x-4x^3)^2] *sqrt(1-x^2) = 0

It proves that arcsin(3x-4x^3) - 3arcsinx = 0