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

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

1.	If `cos^(-1).(6x)/(1 + 9x^(2)) = -(pi)/(2) + 2 tan^(-1) 3x`, then find the values of x
Answer» `cos^(-1).(6x)/(1 + 9 x^(2)) = -(pi)/(2) + 2 tan^(-1) 3x` `(pi)/(2) - sin^(-1).(6x)/(1 + 9x^(2)) = -(pi)/(2) + 2 tan^(-1) 3x` `sin^(-1).(6x)/(1 + 9x^(2)) = pi - 2 tan^(-1) 3x` `sin^(-1).(2 xx 3x)/(1 + (3x)^(2)) = pi - 2 tan^(-1) 3x` It is true when `3x gt 1` or `x gt (1)/(3)` i.e., `x in ((1)/(3), oo)`

`tan^(-1)[(cosx)/(1+sinx)]`is equal to`pi/4-x/2,forx in (-pi/2,(3pi)/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-(3pi)/2,pi/2)`
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