Find the number of values of $x$ satisfying given equation $(1 - x) = cos (2 {sin}^{- 1} x), 0 \leq x \leq \frac{π}{2}$

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Solution

$(1 - x) = c o s [2 ((\frac{π}{2}) - c o s^{- 1} x)] (1 - x) = c o s (π - 2 c o s^{- 1} x) (1 - x) = - c o s (2 c o s^{- 1} x) x - 1 = c o s [c o s^{- 1} (2 x^{2} - 1)] x - 1 = 2 x^{2} - 1 o r 2 x^{2} - x = 0 x (2 x - 1) = 0 x = 0 o r x = (\frac{1}{2})$
Hence, there are 2 values of x which can satisfy the given equation.

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