The solutions of the equation 4cos2x+6sin2x=5 are
x = nπ ± π / 4, nϵZ
x = nπ ± π /3, nϵZ
x = nπ ± π /2, nϵZ
x = nπ ± 2π /3, nϵZ
Given,4cos2x+6sin2x=5⇒4(cos2x+sin2x)+2sin2x=5⇒2sin2x=5−4⇒sin2x=12⇒x=nπ±π4, n∈Z