3 sin2 x - 7 sin x + 2 = 0 का व्यापक हल क्या है?

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AAI ATC Junior Executive 21 Feb 2023 Shift 1 Official Paper
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  1. \(\rm x=\frac{n\pi}{2}+(-1)^n\sin^{-1}\frac{1}{3}\)
  2. \(\rm x={n\pi}+(-1)^n\sin\frac{1}{3}\)
  3. \(\rm x={n\pi}+(-1)^n\sin^{-1}\frac{1}{3}\)
  4. \(\rm x=2{n\pi}+(-1)^n\sin\frac{1}{3}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm x={n\pi}+(-1)^n\sin^{-1}\frac{1}{3}\)
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दिया गया है:

 3 sin2 x - 7 sin x + 2 = 0

अवधारणा:

sin x के व्यापक मान की अवधारणा का प्रयोग कीजिए। 

\(x=\rm n\pi+(-1)^n \theta\)

गणना:

3 sin2 x - 7 sin x + 2 = 0

sin x = y प्रतिस्थापित कीजिए, 

⇒3y2 - 7y + 2 = 0

⇒3y2 - 6y - y + 2 = 0

⇒3y(y - 2) - (y - 2) = 0

⇒(y - 2)(3y - 1) = 0

⇒y = 2 , 1/3

हम जानते हैं कि sin x की परास [-1,1] है, तो 2 इससे अधिक की सीमा में नहीं है,

⇒ sin x = 1/3

⇒ x = sin-1(1/3)

इसलिए व्यापक मान 

\(\rm x={n\pi}+(-1)^n\sin^{-1}\frac{1}{3}\)

अतः विकल्प (3) सही है।

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