If θϵ[0,5π] and rϵR such that 2sinθ=r4−2r2+3 then the maximum number of values of the pair (r,θ) is
2sinΘ=(r2−1)2+2
−2≤2sinΘ≤2
So, possible value of (r2−1)2+2 equals
to 2
Thus, r=±1
Possible values of 2sinθ=2 are π2,5π2,9π2
Then possible pairs are (−1,π2),(−1,5π2),(−1,7π2)(1,π2),(1,5π2),(1,7π2)
Hence, option 'C' is correct.