 Hello and welcome to the session where we are going to understand the following problem today. Sine inverse 1 minus x minus 2, sine inverse x is equal to pi by 2, then x is equal to a 0, half b 1, half c 0, d 1 by 2. Now let us first write the key idea to the problem. Sine inverse x plus cos inverse x is equal to pi by 2, where x belongs to closed interval minus 1 power 1. This is the key idea to our problem. Now let us write the solution. Sine inverse 1 minus x plus cos inverse 1 minus x is equal to pi by 2. Therefore the given equation becomes sine inverse 1 minus x minus 2 sine inverse x is equal to pi by 2, which implies sine inverse 1 minus x minus 2 sine inverse of x is equal to sine inverse or 1 minus x plus cos inverse of 1 minus x. This gets cut off, so we left with minus 2 sine inverse of x is equal to cos inverse or 1 minus x. We will name it as 1. Now let sine inverse x is equal to alpha which implies x is equal to sine alpha. We will name these as 2. Now equation 1 becomes minus 2 alpha is equal to cos inverse of 1 minus x which implies cos 2 alpha is equal to 1 minus x which implies 1 minus 2 sine square alpha is equal to 1 minus x because cos of minus 2 alpha is equal to cos of 2 alpha and cos 2 alpha is equal to 1 minus 2 sine square alpha. Now this becomes putting sine alpha is equal to x in above equation we get 1 minus 2x square minus is equal to 1 minus x which implies 2x square minus x is equal to 0 which implies staying x common so we are left with 2x minus 1 is equal to 0 which implies x is equal to 0 or x is equal to half. Putting x is equal to half the value does not satisfy the equation therefore x is equal to 0 is the answer. Hence the required answer is 3. I hope you understood the problem. Bye and have a nice day.