निम्नलिखित समीकरणों को हल कीजिये -
`sin[2cos^(-1){cot(2tan^(-1)x)}]=0`

Explanation

दिया गया समीकरण है -
` sin [ 2 cos ^(-1) {cot (2 tan ^(-1) x )}]=0`
` rArr sin [ 2 cos^(-1) {cot (tan^(-1).(2x)/(1-x^(2)))}]=0`
` [ :' 2 tan ^(-1) x = tan ^(-1). (2x)/(1-x^(2))]`
` rArr sin [ 2 cos ^(-1) ((1-x^(2))/(2x))]=0`
` [ :' cot^(-1) x = tan ^(-1). 1/x]`
` rArr sin [ 2 cos^(-1) ((1-x^(2))/(2x))]=0`
` [ :' cot ( cot^(-1) x ) = x]`
` rArr [ sin ^(-1) { 2 ((1-x^(2))/(2x))sqrt(1-((1-x^(2))/(2x))^(2))}] = 0 `
` [ :' 2 cos ^(-1) x = sin ^(-1) ( 2x sqrt(1-x^(2)))] `
` rArr 2 ((1-x^(2))/(2x))sqrt(1-((1-x^(2))/(2x))^(2)) = 0 `
` rArr (1-x^(2))/(2x) = 0 `
` sqrt (1- ((1-x^(2))/(2x))^(2))=0`
` rArr 1 - x^(2) = 0 `
`((1-x^(2))/(2x))^(2) = 1`
` rArr x^(2) = 1 " " (1-x^(2))^(2) = 4x^(2)`
` rArr x = pm 1 " " (1-x^(2))^(2) = 4x^(2) `
` (1-x^(2))^(2) = 4x^(2) `
` rArr (1-x^(2))^(2) - (2x)^(2) = 0`
` rArr (1-x^(2))^(2) - (2x)^(2) = 0 `
` rArr (1-x^(2) - 2x) (1-x^(2) + 2x) = 0 `
` rArr (1- 2x - x^(2)) (1+ 2x - x^(2)) = 0 `
` rArr 1 + 2x - x^(2) = 0 `
` 1+ 2x - x^(2) = 0 `
` rArr x^(2) + 2x - 1 = 0 `
` x^(2) - 2x - 1 = 0 `
` rArr x = -1 pm sqrt(2)`
` x = 1 pm sqrt(2)`
अतः ` x = pm 1 , -1 pm sqrt(2) , 1 pm sqrt(2)` अभीष्ट मूल है ।