The number of solutions of sec2 ⁡θ+cosec2⁡ θ+2cosec2 ⁡θ=8,0≤θ≤π/2 is

We have 1sin2 ⁡θcos2 ⁡θ+2sin2⁡ θ=8,sin⁡ θ≠0,cos⁡ θ≠0or  1+2cos2⁡ θ =8sin2⁡ θcos2⁡θor   1+2cos2⁡ θ=8cos2 ⁡θ1−cos2⁡θor    8cos4⁡ θ−6cos2⁡ θ+1=0or   4cos2 ⁡θ−12cos2 ⁡θ−1=0or    cos2⁡ θ=1/4=cos2 ⁡(π/3)or    cos2⁡ θ=1/2=cos2⁡ (π/4)or    θ=nπ±(π/3) or θ=nπ±(π/4),n∈ZHence, for 0≤θ≤π/2,θ=π/3,θ=π/4