Solve the following equations: `sin[2cos^(-1)"{"cot"("2tan^(-1)x"}]"=0`

Options

A. `pm1`
B. `1pm sqrt(2)`
C. `-1pm sqrt(2)`
D. `pm sqrt(2)`

Answer

A, B, C

Explanation

`because sin[2 cos^(-1){cot^(-1){cot(2 tan^(-1)x)}]=0`
`rArr 2 cos^(-1){cot(2 tan^(-1)x)}=n pi, n in I`
`rArr cos^(-1)(cot(2 tan^(-1)x))=(n pi)/(2)`
`rArr cos^(-1)(cot(2 tan^(-1)x))=0, (pi)/(2), pi`
`(because 0 le cos^(-1)x le pi)`
`rArr cot {tan^(-1)((2x)/(1-x^(2)))}=1,0,-1`
`rArr cot{cot^(-1)((1-x^(2))/(2x))}=1,0,-1`
`rArr (1-x^(2))/(2x)=1,0,-1`
`rArr 1-x^(2)=2x,0,-2x`
`rArr x^(2)+2x-1=0, x^(2)-1=0,x^(2)-2x-1=0`
`rArr x=-1 pm sqrt(2), x= pm 1, x=1 pm sqrt(2)`