Question
Download Solution PDFθ के कितने मान समीकरण (sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0 को संतुष्ट करेंगे, जहाँ 0 < θ < ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0, 0 < θ <
गणना:
प्रश्न के अनुसार,
(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0
⇒ (sin2 θ - 4 sin θ + 3) = 0 या (4 - cos2 θ + 4 sin θ) = 0
पहले समीकरण को हल करने पर,
(sin2 θ - 4 sin θ + 3) = 0
⇒ sin2 θ - sin θ - 3 sin θ + 3 = 0
⇒ sin θ (sin θ - 1) - 3 (sin θ - 1) = 0
⇒ (sin θ - 1) (sin θ - 3) = 0
⇒ sin θ = 1 या sin θ = 3
⇒ θ = 900 =
अन्य समीकरण को हल करने पर,
(4 - cos2 θ + 4 sin θ) = 0
⇒ 4 - (1 - sin2 θ) + 4 sin θ = 0
⇒ sin2 θ + 4 sin θ + 3 = 0
⇒ sin2 θ + sin θ + 3 sin θ + 3 = 0
⇒ sin θ (sin θ + 1) + 3 (sin θ + 1) = 0
⇒ (sin θ + 1) (sin θ + 3) = 0
⇒ sin θ = -1 या sin θ = -3
⇒ θ = 2700 =
पुनः, प्रश्न के अनुसार, 0 < θ <
∴ दिया गया समीकरण θ, 0 < θ <
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