θ = (2n+1)π,n  in Z হবে যখন,

Another Explanation (5): প্রশ্ন: \(\theta = (2n+1)\pi, \quad n \in \mathbb{Z}\) হবে কখন? উত্তর: যখন \(\cos \theta = -1\) --- সমাধান: দেওয়া: \(\theta = (2n+1)\pi, \quad n \in \mathbb{Z}\) তাহলে, \(\cos \theta = \cos((2n+1)\pi)\) জানা: \(\cos((2n+1)\pi) = -1\) অর্থাৎ, \(\cos \theta = -1\) অতএব, \(\theta = (2n+1)\pi, \quad n \in \mathbb{Z}\) হলে, \(\cos \theta = -1\) **উত্তর:** \(\boxed{\cos \theta = -1}\)