Question

Solve the equation for x: 

sin^-1x + sin^-1(1 - x) = cos^-1x, x ≠ 0 

Answer 1

sin^-1x + sin^-1(1 - x) = cos^-1x

`=>"sin"^-1{"x"sqrt(1-(1-"x")^2) + (1-"x")sqrt(1-"x"^2)} = "sin"^-1sqrt(1-"x"^2)`

`[∵ "sin"^-1"x" + "sin"^-1"y" = "sin"^-1{"x"sqrt(1-"y"^2) + "y"sqrt(1-"x"^2)} "and"  "cos"^-1"x" = "sin"^-1 sqrt(1-"x"^2)]`

`=> "x"sqrt(1-1+2"x" - "x"^2) + sqrt(1-"x"^2) - "x"sqrt(1-"x"^2) = sqrt(1-"x"^2)`

 

`=> "x"sqrt(2"x"-"x"^2) - "x" sqrt(1-"x"^2) = 0`   

`=> "x"(sqrt(2"x"-"x"^2) - sqrt(1-"x"^2))=0`

`=>"x" = 0 , sqrt(2"x" - "x"^2) - sqrt(1-"x"^2) = 0`  

`=> x=0 , sqrt(2"x" - "x"^2) = sqrt(1-"x"^2)`

 

Now, `sqrt(2"x" - "x"^2) = sqrt(1-"x"^2)`

 

Squaring both sides, we obtain

2x - x² = 1 - x² 

⇒ 2x = 1

⇒ x = `1/2`

Hence, x = 0 and x = `1/2`