Now we will put Now, we will put x= siny in the given equation, and we get,

⟹ 1 – siny = cos2y as

⟹ 1 - cos2y = sin y

⟹ 2sin²y = sin y

⟹ siny(2siny - 1) = 0

⟹ siny = 0 or 1/2

∴ x = 0 or 1/2

Now, if we put then we will see that,

L.H.S. =

R.H.S

Hence, is not the solution of the given equation.

Thus, x =