Sin^-1(1-x)-2sin^-1(x)=(π÷2) then find the value of x

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Arun · Answer

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 solving 1-x = 1-2 x² 2 x² -x = 0 x = 0 and x = ½ But for x = 1/2 , LHS is not equal to RHS. Hence x = 0 is the solution. Regards Arun (askIITians forum expert)

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Sin^-1(1-x)-2sin^-1(x)=(π÷2) then find the value of x

Sin^-1(1-x)-2sin^-1(x)=(π÷2) then find the value of x

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