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If cos^(-1)(6x)/(1+9x^2)=-pi/2+tan^(-1)3...

If `cos^(-1)(6x)/(1+9x^2)=-pi/2+tan^(-1)3x ,` then find the value of`xdot`

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`cos^(-1)((6x)/(1+9x^2))=-pi/2+2tan^(-1)3x`
`pi/2-sin^(-1)((6x)/(1+9x^2))=-pi/2+2tan^(-1)x`
`pi-sin^(-1)((6x)/(1+9x^2))=2tan^(-1)3x`
`pi-2tan^(-1)3x=sin^(-1)((2(3x))/(1+(3x)^2))`
`pi-2tan^(-1)3x=sin^(-1)((2(3x))/(1+(3x)^2))`
When `3x>=1,x>1/3`
`x in(1/3,oo)`.
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