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If logcos x sin x>=2 and 0<=x<=3π then sin x is in the interval





7 years ago


Answers : (1)


logcos xsin x >=2

sin x >= cos2x

sin x >= 1 - sin2x

sin2x + sin x - 1 >= 0

[sin x - (51/2-1) / 2] * [sin x - (-51/2-1) / 2] >= 0

now, sin x > -1

therefore [sin x - (-51/2-1)/2] > 0       { 51/2 ≈ 2        therefore    -(-51/2-1)/2 ≈ 1.5}

therefore [sin x - (51/2-1) / 2] should be > 0

sin x > (51/2-1) / 2]

and sin x lies between [-1,1]


7 years ago

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