First equation is 4sin2θcos2θ=2cos2θ ⇒cos2θ=0 or sin2θ=21=(21)2=(sin)24π ⇒θ=(2n+1)2π,n∈I or θ=nπ±4π,n∈I
Second equation is not satisfied by θ=(2n+1)2π,n∈I but satisfied by θ=nπ±4π,n∈I So θ=nπ±4π,n∈I ∴ In [2−3π,34π] , the values of θ are 4−5π,4−3π,4−π,4π,43π,45π