The number of solutions of the equation cot 2 ⁡ sin ⁡ x + 3 = 1 in 0,3 π is equal to

Step-by-step solution

cot 2 ⁡ sin ⁡ x + 3 = 1 = cot 2 ⁡ π 4 ⇒ sin ⁡ x + 3 = n π ± π 4 Also, 2 ≤ sin ⁡ x + 3 ≤ 4 ⇒ sin ⁡ x + 3 = π - π 4 , π + π 4 ⇒ sin ⁡ x = 3 π 4 - 3 or 5 π 4 - 3