If f(θ)=sin⁡θcos⁡θsin⁡θcos⁡θsin⁡θcos⁡θcos⁡θsin⁡θsin⁡θ, then

f(θ)=sin3⁡θ+cos3⁡θ−cos⁡θsin⁡θ(sin⁡θ+cos⁡θ)       =(sin⁡θ+cos⁡θ)3−4sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)       =(sin⁡θ+cos⁡θ)[1−sin⁡2θ]Now, f(θ)=0⇒ tan⁡θ=−1 or sin⁡2θ=1⇒ f(θ)=0 has 2real solutions in [0,π]Also, f(θ)1−sin⁡2θ=sin⁡θ+cos⁡θ∈[−2,2]