निम्नलिखित समीकरणों को हल कीजिये -
(i) `sin^(-1). (3x)/5 + sin^(-1). (4x)/5 = sin^(-1) x `

Explanation

दिया गया समीकरण है -
` sin ^(-1). (3x)/5 + sin^(-1). (4x)/5 = sin^(-1) x`
` rArr sin^(-1) [ (3x)/5 sqrt(1-((4x)/5)^(2)) +(4x)/5 sqrt(1-((3x)/5)^(2))] = sin^(-1) x`
` [ :' sin^(-1) y = sin ^(-1) [ x sqrt(1-y^(2))+ysqrt(1-x^(2))]`
` rArr sin^(-1) [ (3x)/5 xx sqrt(1-(16x^(2))/25)+(4x)/5 sqrt(1-(9x^(2))/25)] = sin^(-1) x`
` (3xsqrt(25 - 16x^(2)))/25 + (4xsqrt(25-9x^(2)))/25 = x `
` rArr 3x sqrt(25 - 16x^(2)) + 4x sqrt(25 - 9x^(2)) = 25 x `
` rArr x [ 3 sqrt(25 - 16x^(2))+4x sqrt(25 - 9x^(2))] -25 x = 0 `
` rArr " " x = 0`
`3 sqrt(25 - 16x^(2)) +4sqrt(25 - 9x^(2))= 25 `
` rArr 4 sqrt(25 - 9x^(2)) = 25 - 3 sqrt(25 - 16x^(2))`
` rArr 16 (25 - 9x^(2)) = 625 + 9 (25 - 16x^(2))`
` rArr 16 (25 - 9x^(2)) = 625 + 9 (25 - 16x^(2))`
` - 150 sqrt(25- 16x^(2))`
[दोनों पक्षों का वर्ग करने पर ]
` rArr 150 sqrt(25 - 16x^(2)) = 450 `
` rArr 25 - 16x^(2) = 3^(2)`
` rArr 16x^(2) = 16`
` rArr x^(2) = 1`
` rArr x = pm 1 `
अतः ` x= 0 , 1,-1` समीकरण के मूल है ।