We are given with equation:

√(1 + cos 2x) = √2 sin^-1(sin x) …(i)

Where -π ≤ x ≤ π

We need to find the number of real solutions of the given equation.

Using trigonometric identity,

cos 2x = cos² x – sin² x

⇒ cos 2x = cos² x – (1 – cos² x) [∵, sin² x + cos² x = 1 ⇒ sin² x = 1 – cos² x]

⇒ cos 2x = cos² x – 1 + cos² x

⇒ cos 2x = 2 cos² x – 1

⇒ 1 + cos 2x = 2 cos² x

Substituting the value of (1 + cos 2x) in equation (i),

√(2 cos² x) = √2 sin^-1(sin x)

⇒ √2 |cos x| = √2 sin^-1(sin x)

√2 will get cancelled from each sides,

⇒ |cos x| = sin^-1(sin x)

Take interval :

|cos x| is positive in interval , hence |cos x| = cos x.

And, sin x is also positive in interval , hence sin^-1(sin x) = x.

So, |cos x| = sin^-1(sin x)

⇒ cos x = x

If we draw y = cos x and y = x on the same graph, we will notice that they intersect at one point, thus giving us 1 solution.

∴, There is 1 solution of the given equation in interval .

Take interval :

|cos x| is negative in interval , hence |cos x| = -cos x.

And, sin x is also negative in interval , hence sin^-1(sin (π + x)) = π + x.

So, |cos x| = sin^-1(sin x)

⇒ -cos x = π + x

⇒ cos x = -π – x

If we draw y = cos x and y = -π – x on the same graph, we will notice that they intersect at one point, thus giving us 1 solution.

∴, There is 1 solution of the given equation in interval .

Take interval :

|cos x| is negative in interval , hence |cos x| = -cos x.

And, sin x is positive in interval , hence sin^-1(sin (-π – x)) = -π – x.

So, |cos x| = sin^-1(sin x)

⇒ -cos x = -π – x

⇒ -cos x = -(π + x)

⇒ cos x = π + x

If we draw y = cos x and y = π + x on the same graph, we will notice that they doesn’t intersect at any point, thus giving us no solution.

∴, There is 0 solution of the given equation in interval .

Hence, we get 2 solutions of the given equation in interval [-π, π].