If 3 ⁻¹ ( 2 x 1+x^2 )-4 ⁻¹ ( 1-x^2 1+x^2 )+2 ⁻¹ ( 2 x 1-x^2 )= 3 , then the value of x=

Step-by-step solution

Given the equation 3 ⁻¹ ( 2x 1+x^2 )-4 ⁻¹ ( 1-x^2 1+x^2 )+2 ⁻¹ ( 2x 1-x^2 )= 3 . Using standard identities for inverse trigonometric functions, the relevant ranges for x > 0 are considered. For 0 ⁻¹ ( 2x 1+x^2 )=2 ⁻¹x , ⁻¹ ( 1-x^2 1+x^2 )=2 ⁻¹x , ⁻¹ ( 2x 1-x^2 )=2 ⁻¹x . Substituting into the equation: 3(2 ⁻¹x)-4(2 ⁻¹x)+2(2 ⁻¹x)= 3 2 ⁻¹x= 3 ⁻¹x= 6 x= ( 6 )= 1 3 . This value satisfies 0 For x = 1 : The term ⁻¹ ( 2 0 ) is undefined, so no solution. For x > 1 : ⁻¹ ( 2x 1+x^2 )= -2 ⁻¹x , ⁻¹ ( 1-x^2 1+x^2 )=2 ⁻¹x , ⁻¹ (