Sol. ∴ Given equation is

1 + (sin x – cos x) sin = 2 cos 2 .... (1)

⇒ 1 + (sin x – cos x) = 1 + cos 5x

⇒ cos 5x + cos = 0

⇒ 2 cos cos = 0

⇒ cos = 0, cos = 0 ..... (2)

whose roots are equal, respectively, to x = +

or x = + , n ∈ I

Period of (2) are and π .

From the obtained solutions of the equation belonging to the interval [0, 2 π ] the number and π + satisfy the inequality sin 6x < 0, all the solutions of the problem can be obtained by adding number, which are multiplies of 2 π , to each roots obtained.

x = n π + , n ∈ I