Cos(2 cos−1 x)+ 3cos(cos−1 x)− 4 = 0

x = 1

cos(2 cos^-1 x) + 3 cos(cos^-1 x) - 4 = 0

cos(2 arccos x) + 3 cos(arccos x) - 4 = 0 , x € [-1 , 1]

*** cos(2t) = cos(t)^2 - sin(t)^2 ***

cos(arccos x)^2 - sin(arccos x)^2 + 3cos(arccos x) - 4 = 0

*** cos(arccos t) = t ***

x^2 - sin(arccos x)^2 + 3x - 4 = 0

*** sin(t)^2 = 1 - cos(t^2) ***

x^2 - (1 - cos(arccos x)^2) + 3x - 4 = 0

x^2 - (1 - x^2) + 3x - 4 = 0

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

2x^2 - 5 + 3x = 0

2x^2 + 3x - 5 = 0

2x^2 + 5x - 2x - 5 = 0

x(2x + 5) - (2x + 5) = 0

(2x + 5)(x - 1) = 0

2x + 5 = 0 | x - 1 = 0

2x = -5 | x = 1

x = -5/2 , x € [-1 , 1] | x = 1

x = 1

should have been evaluated with one odd value and one even value. the expressions have equivalent values when . when , the first expression has a value of 3 and the second

step-by-step explanation: