Am I reading it right?
sin(cos^(-1)(x^2/(4 - x^2)))
Solve for x:
sqrt(1 - x^4/(4 - x^2)^2) = 0
Square both sides:
1 - x^4/(4 - x^2)^2 = 0
Bring 1 - x^4/(4 - x^2)^2 together using the common denominator (x^2 - 4)^2:
-(8 (x^2 - 2))/(x^2 - 4)^2 = 0
Divide both sides by -8:
(x^2 - 2)/(x^2 - 4)^2 = 0
Multiply both sides by (x^2 - 4)^2:
x^2 - 2 = 0
Add 2 to both sides:
x^2 = 2
Take the square root of both sides:
Answer: |x = sqrt(2) or x = -sqrt(2)
\(sin\left(cos^{-1}\left(\frac{x^2}{4-x^2}\right)\right)\)
We are looking for the sin of an angle (theta) with a cosine of x^2 / ( 4 - x^2).....so we have...
sin (theta) = √ [ 1 - cos^2 (theta) ] = [ 1 - [ (x^2)/(4-x^2)]^2 ] = √ [x^4 - 8x^2 + 16 - x^4] / (4 - x^2) =
√ {16 - 8x^2) / (4- x^2) = √ [ 4 (4 - 2x^2) ] / (4 - x^2) = 2√(4 - 2x^2) / (4 - x^2)