x = n π + (– 1)n , y = 2n π ±

Let 4 sin x = u and 31/cos y = v. Then, the given system of equations reduces to

u + v = 11 ..... (i)

5u 2 – 2v = 2 ..... (ii)

Eliminating v between (i) and (ii), we get

5u 2 – 22 + 2u = 2

⇒ 5u2 – 22 + 2u = 2

⇒ (5u + 12) (u – 2) = 0

⇒ u – 2 = 0 [  u = 4 sinx > 0 for all x ∴ 5u + 12 > 0]

⇒ u = 2

⇒ 4 sinx = 2

⇒ 2 2 sinx = 21

⇒ 2 sinx = 1

⇒ sinx =

⇒ sinx =

⇒ sinx = sin

⇒ x = n π + (–1)n ; n ∈ Z

putting the value of u in (i), we get

v = 9

⇒ 31/cosy = 3 2

⇒ = 2

⇒ cos y =

⇒ cos y = cos

⇒ y = 2m π ± ; n ∈ Z

Hence, x = n π + (–1)n and y = 2 m π ± , where m, n ∈ Z

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