The number of values of x in the interval [0,3π] satisfying the equation 2sin2x+5sinx−3=0 is
2
4
6
1
We have (sinx+3)(2sinx−1)=0
⇒ sinx=−3 which is not possible
or sinx=1/2⇒x=π/6,5π/6,13π/6,17π/6 as x∈[0,3π]