Question
Download Solution PDF0 < θ < π/2 শর্তে, (sin²θ - 4sinθ + 3)(4 - cos²θ + 4sinθ) = 0 সমীকরণটি কতগুলি θ-এর মান সিদ্ধ করবে?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFপ্রদত্ত:
(sin²θ - 4sinθ + 3)(4 - cos²θ + 4sinθ) = 0, 0 < θ < π/2
গণনা:
প্রশ্নানুসারে
(sin²θ - 4sinθ + 3)(4 - cos²θ + 4sinθ) = 0
=> (sin²θ - 4sinθ + 3) = 0 অথবা (4 - cos²θ + 4sinθ) = 0
প্রথম সমীকরণটি সমাধান করলে:
(sin²θ - 4sinθ + 3) = 0
=> sin²θ - sinθ - 3sinθ + 3 = 0
=> sinθ(sinθ - 1) - 3(sinθ - 1) = 0
=> (sinθ - 1)(sinθ - 3) = 0
=> sinθ = 1 অথবা sinθ = 3
=> θ = 90° = π/2 (∵ -1 < sinθ < 1, sinθ = 3 বাদ দিচ্ছি)
অন্য সমীকরণটি সমাধান করলে:
(4 - cos²θ + 4sinθ) = 0
=> 4 - (1 - sin²θ) + 4sinθ = 0
=> sin²θ + 4sinθ + 3 = 0
=> sin²θ + sinθ + 3sinθ + 3 = 0
=> sinθ(sinθ + 1) + 3(sinθ + 1) = 0
=> (sinθ + 1)(sinθ + 3) = 0
=> sinθ = -1 অথবা sinθ = -3
=> θ = 270° = 3π/2 (∵ -1 < sinθ < 1, sinθ = -3 বাদ দিচ্ছি)
আবার, প্রশ্নানুসারে 0 < θ < π/2
∴ প্রদত্ত সমীকরণটি 0 < θ < π/2 শর্তে কোনও θ-এর মানের জন্য সিদ্ধ হয় না।
Last updated on Jun 26, 2025
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