Question

The number of solutions of the pair of equation 2 sin² ø—cos2ø = 0, 2 cos²ø - 3 sin ø = 0 in the interval [0,2π]

(A) Zero

(B) One

(C) Two

(D) Four

Answer 1

(C) Two.

Explanation:

2 sin^2 ø – cos^2 ø = 0
⇒ sin^2 ø = cos^2 ø
⇒ sin ø = cos ø
⇒ ø = π/4, 5π/4

Now, 2 cos^2 ø – 3 sin ø = 0
⇒ 2 (1 – sin^2 ø) – 3 sin ø = 0
⇒ 2 – 2 sin^2 ø – 3 sin ø = 0
⇒ 2 – 2 (1 – cos ø) – 3 sin ø = 0
⇒ 2 – 2 + 2 cos ø – 3 sin ø = 0
⇒ 2 cos ø – 3 sin ø = 0

At ø = π/4, 2 cos (π/4) – 3 sin (π/4) = 2(1/√2) – 3(1/√2) = √2 – 3√2 = -2√2 ≠ 0

At ø = 5π/4, 2 cos (5π/4) – 3 sin (5π/4) = 2(-1/√2) – 3(-1/√2) = -2√2 + 3√2 = √2 ≠ 0

Therefore, the given pair of equations has no solution in the interval [0, 2π].

Answer 2

Correct option is (C) Two

Given, \(2\sin^2 \theta - \cos 2\theta = 0\) and \(2\cos^2 \theta - 3\sin\theta = 0\)

We have,

Therefore, the number of solutions = 2