The correct option is A θ=nπ,nεI
2cos2θx3+2x+sin 2θ=(x2+px+1)(2 cos2θx+sin 2θ)
On comparing coefficients of x and x2, then
2=p sin 2θ+2 cos2θ⇒2 sin2θ=p sin 2θor p=tanθ ...(i)and 0=sin 2θ+2p cos2θ∴p=−tan θ ...(ii)
From Eqs. (i) and (ii),
tanθ=−tanθ2 tan θ=0tanθ=0θ=nπ,nεI.