If log cos x ⁡ sin x ≥ 2 and 0 ≤ x ≤ 3 π , then the value of sin x lies in the interval

Step-by-step solution

Since, sin x > 0 , cos x > 0 , cos x ≠ 1 So, sin x ≤ cos 2 ⁡ x ⇒ sin 2 ⁡ x + sin x - 1 ≤ 0 ⇒ sin x + 1 2 2 - 5 4 ≤ 0 ⇒ | s i n x + 1 2 | ≤ 5 2 ∴ sin x + 1 2 ∈ [ − 5 2 , 5 2 ] (As, sin x > 0 ) ∴ 0< sin x ≤ 5 − 1 2