If 3 sin ⁡ x + cos ⁡ x - 2 = y - 1 2 for 0 ≤ x ≤ 8 π, then the number of values of the pair x , y is equal to

Question

If 3 sin ⁡ x + cos ⁡ x - 2 = y - 1 2 for 0 ≤ x ≤ 8 π, then the number of values of the pair x , y is equal to,

MARKS App · Accepted Answer

max . ⁡ 3 sin ⁡ x + cos ⁡ x = 3 2 + 1 2 = 2 and y - 1 2 ≥ 0 So, LHS ≤ 0 and RHS ≥ 0 So, the possibility of equality is only when LHS = RHS = 0 RHS = 0 ⇒ y = 1 LHS = 0 ⇒ 3 sin ⁡ x + cos ⁡ x = 2 ⇒ 3 2 sin ⁡ x + 1 2 cos ⁡ x = 1 ⇒ cos ⁡ x - π 3 = 1 ⇒ x - π 3 = 2 n π ⇒ x = 2 n π + π 3 0 ≤ 2 n π + π 3 ≤ 8 π ⇒ n = 0 , 1 , 2 , 3 ⇒ There are four values of x and hence four pairs of x , y

Search any question & find its solution

Looking for more such questions to practice?