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Answer 1

Given :  The equation sin²x + asinx + b = 0 , x ∈ (0, π) has exactly 3 distinct solutions

To Find :  b

Solution:

sin²x + asinx + b = 0

=> sinx  = ( - a  ± √a² - 4b )/2

Equation has exactly 3 distinct solution hence one solution must be

1  and other must by between 0 to 1

( - a  ± √a² - 4b )/2 = 1

=>  - a  ± √a² - 4b = 2

=>   ± √a² - 4b = 2 + a

=> a² - 4b = 4 + a²  + 4a

=> 4a + 4b = - 4

=> a + b = - 1

0  < - a  ± √a² - 4b   <  1

sinx  = ( - a  ± √a² - 4b )/2

a + b = - 1

=   (b + 1 ±  (b - 1))/2

= 2b/2  =  b  or  1

Hence

 0  <  b  <  1    

Hence b  ∈ (0 , 1)

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