Ans.

Hint : For (i) to (iii)

The given equation is possible if

sin (1 – x) ≥ 0 and cos x ≥ 0

sin (1 – x) = cos x

sin (1 – x) = sin

1 – x = n π + (–1) n , n ∈ I

(i)

Sol. When n is odd integer

Let n = 2k + 1, k ∈ I

1 – x = n π + (–1) n

x =

∴ if k ≥ 0 then x < 0

But given x > 0, so k < 0

Let k = –1

x =

But < 0 possible in second quadrant

Let k = –2

x = is fourth quadrant

∴ cos x > 0 and

sin(1 – x) = sin = sin

= > 0

∴ Smallest positive root is x =

(ii)

Sol. When n is even Then let n = 2k

1 – x = 2k π + – x which is impossible ∴ No solution.

(iii)

Sol. Clearly from Q. 19

No. of solution in [0, π ] is 1.