x = cos^(-1)(-1/2) = 2π/3

If x = cos^(-1)(1/2), then x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.

If x = cos^(-1)(1/2), then cos(x) = 1/2, which corresponds to x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.

Since x = cos^(-1)(1/2) = π/3, then sin^(-1)(1/2) = π/6.

Since cos^(-1)(1/2) = π/3, we have sin^(-1)(√(1 - (1/2)^2)) = sin^(-1)(√(3/4)) = π/3.

cos^(-1)(1/2) = π/3.

If x = cos^(-1)(1/2), then x = π/3, thus sin(x) = sin(π/3) = √3/2.

Using the identity sin(x) = sqrt(1 - cos^2(x)), we have sin(x) = sqrt(1 - (1/2)^2) = √3/2.

sin^(-1)(-1) corresponds to the angle -π/2.

sin^(-1)(-1) corresponds to the angle -π/2.

sin^(-1)(-1/2) = -π/6, since sin(-π/6) = -1/2.

sin^(-1)(-1/2) = -π/6, since sin(-π/6) = -1/2.

If x = sin^(-1)(1/2), then x = π/6. Therefore, cos(x) = cos(π/6) = √3/2.

If x = sin^(-1)(1/2), then x = π/6. Therefore, cos(x) = cos(π/6) = √3/2.

Using the identity cos(x) = √(1 - sin^2(x)), we find cos(sin^(-1)(1/3)) = √(1 - (1/3)^2) = √(8)/3.

Using the identity cos^(-1)(√(1 - sin^2(x))) = π/2 - x, we find that cos^(-1)(√(1 - (1/3)^2)) = π/2 - sin^(-1)(1/3).

If x = sin^(-1)(1/√2), then sin(x) = 1/√2. Therefore, cos(x) = √(1 - sin^2(x)) = √(1 - (1/√2)^2) = √(1/2) = √2/2.

Since x = sin^(-1)(1/√2) = π/4, then cos^(-1)(x) = π/2 - π/4 = π/4.

Using the identity cos^2(x) + sin^2(x) = 1, we find cos(x) = √(1 - (3/5)^2) = 4/5.

Using the identity cos(x) = sqrt(1 - sin^2(x)), we have cos(x) = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5.

tan^(-1)(1) = π/4.

tan^(-1)(1) = π/4, since tan(π/4) = 1.

tan^(-1)(1/√3) = π/6, since tan(π/6) = 1/√3.

x = tan^(-1)(√3) = π/3, thus sin^(-1)(x) = sin^(-1)(√3/2) = π/3.

Since tan^(-1)(√3) = π/3, then 2x = 2π/3 and sin(2x) = sin(2π/3) = √3/2.

This is a perfect square: (x + 1)^2 = 0 => x = -1.

This is a perfect square: (x - 3)^2 = 0, so x = 3.

x^3 = 8 => x = 2

Taking the cube root of both sides, x = 8^(1/3) = 2.

Taking the fourth root, x = 81^(1/4) = 3.