If 4sin2x–8sinx+3≤0, then the solution set for x≤0≤x≤2π is
0,π6
0,5π6
5π6,2π
π6,5π6
The explanation for the correct option:
⇒4sin2x–8sinx+3≤0⇒4sin2x−2sinx−6sinx+3⩽0⇒2sinx(2sinx−1)−3(2sinx−1)⩽0⇒(2sinx−1)(2sinx−3)⩽0
Hence,
sinx∈12,32
but we knowsinx∈[−1,1]
So required values of sinx lies in 12,1
∴x∈π6,5π6
for 0⩽x⩽2π
Hence, the correct answer is option (D).