Solve for x: sin–1 (1 – x) – 2 sin–1 (x) = π/2, -1 denotes inverse function?

A = sin^-1(x-1)

B = sin^-1(x)

Take cosine,

cos(A-2B) = 0

cosA cos2B + sinA sin2B = 0

sqrt[1-(x-1)^2] [1-2x^2] + (x-1)(2x)sqrt(1-x^2) = 0

sqrt(x)[sqrt(2-x)(1-2x^2) + 2sqrt(x)(x-1)sqrt(1-x^2)] = 0

So, x = 0 is a solution.

Since f(x) = A - 2B, f'(x) = -1/sqrt[1-(1-x)^2] - 2/sqrt(1-x^2) < 0, f(x) is decreasing. Thus, x = 0 is the only solution.

Solve for x: sin–1 (1 – x) – 2 sin–1 (x) = π/2, -1 denotes inverse function?

sin⁻¹(1-x)-2sin⁻¹x=∏/2

sin⁻¹(1-x)=(∏/2)+2sin⁻¹x

1-x =sinx[(∏/2)+2sin��¹x]

=cos(2sin⁻¹x)

Put Ѳ=sin⁻¹x

sinѲ=x

1-x=cos2Ѳ

=1-2sin²Ѳ

1-x =1-2x²

2x²-x+1-1=0

X(2x-1)=0

X=0 or 2x-1=0

2x=1

X=1/2

Therefore x=0 and ½

Since ½ is not satisfy the equation, x=0

sin(a - b) = sin(a)cos(b) - sin(b)cos(a)

cos(arcsin(t)) = sqrt(1 - t^2)

arcsin(1 - x) - 2 * arcsin(x) = pi/2

sin(arcsin(1 - x) - 2 * arcsin(x)) = sin(pi/2)

sin(arcsin(1 - x)) * cos(2 * arcsin(x)) - sin(2 * arcsin(x)) * cos(arcsin(1 - x)) = 1

(1 - x) * (cos(arcsin(x))^2 - sin(arcsin(x))^2) - 2 * sin(arcsin(x)) * cos(arcsin(x)) * cos(arcsin(1 - x)) = 1

(1 - x) * (1 - x^2 - x^2) - 2 * x * sqrt(1 - x^2) * sqrt(1 - (1 - x)^2) = 1

(1 - x) * (1 - 2x^2) - 2x * sqrt(1 - x^2) * sqrt(1 - (1 - x)^2) = 1

(1 - x) * (1 - 2x^2) - 1 = 2x * sqrt(1 - x^2) * sqrt(1 - (1 - x)^2)

1 - x - 2x^2 + 2x^3 - 1 = 2x * sqrt(1 - x^2) * sqrt(1 - (1 - x)^2)

2x^3 - 2x^2 - x = 2x * sqrt(1 - x^2) * sqrt(1 - 1 + 2x - x^2)

Now we can already see that one solution is x = 0, so let's eliminate that solution right now for the sake of simplicity:

2x^2 - 2x - 1 = 2 * sqrt(1 - x^2) * sqrt(2x - x^2)

(2x^2 - 2x - 1)^2 = 4 * (1 - x^2) * (2x - x^2)

4x^4 - 4x^3 - 2x^2 - 4x^3 + 4x^2 + 2x - 2x^2 + 2x + 1 = 4 * (2x - x^2 - 2x^3 + x^4)

4x^4 - 8x^3 + 0x^2 + 4x + 1 = 8x - 4x^2 - 8x^3 + 4x^4

4x + 1 = 8x - 4x^2

4x^2 + 4x - 8x + 1 = 0

4x^2 - 4x + 1 = 0

(2x - 1)^2 = 0

(2x - 1) = 0

2x - 1 = 0

2x = 1

x = 1/2

However, since we squared our equation, x = 1/2 may be extraneous. But now we have 2 solutions:

x = 0 , 1/2

Test:

arcsin(1 - 0) - 2 * arcsin(0) =>

arcsin(1) - 2 * arcsin(0) =>

pi/2 - 2 * 0 =>

pi/2 - 0 =>

pi/2

arcsin(1 - 1/2) - 2 * arcsin(1/2) =>

arcsin(1/2) - 2 * arcsin(1/2) =>

pi/6 - 2 * pi/6 =>

-pi/6

x = 0 is the only viable answer

1) Given: sin⁻¹(1-x) - 2sin⁻¹(x) = π/2

2) ==> sin⁻¹(1-x) = π/2 + 2sin⁻¹(x)

3) Taking sine both sides, the above implies,

(1 - x) = sin{π/2 + 2sin⁻¹(x)} = cos{2sin⁻¹(x)}

4) Considering, sin⁻¹(x) = θ, cos{2sin⁻¹(x)} = cos(2θ) = 1 - 2sin²θ = 1 - 2(sinθ)²

==> = 1 - 2[sin{sin⁻¹(x)}]² = 1 - 2x² [Since, sin{sin⁻¹(x)} = x]

5) Thus, the above simplifies to, 1 - x = 1 - 2x²

==> 2x² - x = 0

Solving, either x = 0 or x = 1/2

But x = 1/2 does not satisfy the given equation.

Hence, the only solution is: x = 0

y=(x+1)^2 Now switch x and y. x = (y+1)^2 Now, solve for y. First, take the squar root of both sides. SqRt x = y + 1 Subtract 1 from both sides. SqRt x - 1 = y

Sin-1x

arcsin(1 - x) - 2arcsin(x) = pi/2

Substitute x = 0:

arcsin(1) - 2arcsin(0) = pi/2

pi/2 - 2 * 0 = pi/2

pi/2 = pi/2

Hence proven.