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Sin^-1(1-x)-2sin^-1(x)=(π÷2) then find the value of x
Dear Vaibhav Sin^-1(1-x) = (pi/2) + 2 sin^-1(x)(1-x) = sin[pi/2 +(2sin^-1(x)]1-x = cos (2 sin^-1(x))On changing sin^-1 x to cos ^-1 x, and further solving1-x = 1-2 x²2 x² -x = 0x = 0 and x = ½But for x = 1/2 , LHS is not equal to RHS.Hence x = 0 is the solution. RegardsArun (askIITians forum expert)
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