The number of values of x in the interval [0,5π] satisfying the equation 3sin2x−7sinx+2=0 is
3sin2x−7sinx+2=0
or (sinx−2)(3sinx−1)=0
⇒sinx=13=sinα (say)(sinx=2 is not possible)
There exists two values in(0,π),(2π,3π) and (4π,5π).
Hence there are six solutions