The equation 2cos2x2sin2x=x2+1x2,0≤x≤π2has
one real solution
no solution
more than one real solution
two solutions
Since x2+x−2≥2 and 2cos2x2sin2x≤2
⇒2cos2x2sin2x=2 ⇒cos2x2sin2x=1 ⇒cosx2=sinx=1 which is not possible for any choice of x.