(i) The given equation is

⇒ sin [2 cos –1 {cot (2 tan –1 x)}] = 0

⇒ sin [2 cos –1 ] = 0

⇒ sin = 0

⇒ sin = 0

⇒ sin cos –1 = 0

⇒ sin sin –1 = 0

⇒ 4x 4 – (x 4 – 4x 2 + 1) 2 = 0

⇒ (x 4 –2x 2 + 1) (x 4 – 6x 2 + 1) = 0

⇒ either x 4 –2x 2 + 1 or x 4 – 6x 2 + 1 = 0

⇒ (x 2 –1) 2 = 0 or (x 2 –3) 2 = 8

⇒ x = ± 1 or x 2 = 3 ± 2

∴ Solution x = ± 1, ± (1± )

(ii) We have (sin –1 x) 3 + (cos –1 x) 3 = απ 3 ⇒ (sin

–1 x + cos –1 x) {(sin –1 x) 2

+ (cos –1 x) 2 – sin –1 x cos –1 x} = απ 3 ⇒ {(sin

–1 x) 2 + (cos –1 x) 2 – sin –1 x cos –1 x} = απ 3 ⇒ (sin

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

⇒ (sin –1 x) 2 + – sin –1 x = 2 π 2 α

⇒ 3 (sin –1 x) 2 – sin –1 x + = 2 π 2 α

⇒ (sin – 1 x) 2 – sin –1 x + – = 0

⇒ – + – = 0

⇒ + (1 – 32 α ) = 0

⇒ =

Hence the given equation has no solution if R.H.S. is negative or α – < 0

i.e. if α <