If logcos x sin x>=2 and 0<=x<=3π then sin x is in the interval(a)[(51/2-1)/2,1](b)(0,51/2-1)/2](c)[0,1/2](d)none

log_{cos x}sin x >=2

sin x >= cos²x

sin x >= 1 - sin²x

sin²x + sin x - 1 >= 0

[sin x - (5^1/2-1) / 2] * [sin x - (-5^1/2-1) / 2] >= 0

now, sin x > -1

therefore [sin x - (-5^1/2-1)/2] > 0       { 5^1/2 ≈ 2        therefore    -(-5^1/2-1)/2 ≈ 1.5}

therefore [sin x - (5^1/2-1) / 2] should be > 0

sin x > (5^1/2-1) / 2]

and sin x lies between [-1,1]

(a)[(5^1/2-1)/2,1]