Let cosθ=t
Then, (log2t)2+log4/t(16t)=2
⇒(log2t)2+log2(16t)log2(4t)=2 (Using base change property)
⇒(log2t)2+4+log2t2−log2t=2
Let log2t=z
z2+4+z2−z=2
⇒z(z2−2z−3)=0
⇒z=0,−1,3
⇒t=1,12,8
So, cosθ=12 or cosθ=1
⇒θ=2nπ±π3 or θ=2mπ ; n,m∈I
Hence, the number of solutions in [0,2π) is 3(θ=π3,5π3,0)