the number of real solutions of equation is:
sqrt(1 + cos2x) = sqr(2).sin^-1(sinx)
x belongs to [-pi, pi]
options: 0, 1, 2, infinite

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\sqrt{1 + \cos 2 x} = \sqrt{2} \sin^{- 1} (\sin x), - π \leq x \leq π

\sqrt{1 + c o s 2 x} = \sqrt{2} s i n^{- 1} (s i n x), - π < x \leq π

\sqrt{1 - c o s 2 x} = \sqrt{2} s i n^{- 1} (s i n x), - π < x \leq π

sin (2^{x}) = π^{x} + π^{- x}

sin 2 x + cos 2 x = 0

π < x < 2 π

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