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प्रश्न
If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
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उत्तर
\[\text{ Let } f\left( x \right) = 12\sin x - 9 \sin^2 x \]
\[ = - \left( 9 \sin^2 x - 12 \sin x \right) \]
\[ = - \left[ \left( 3\sin x \right)^2 - 2 . 3 \sin x . 2 + 2^2 - 4 \right]\]
\[ = - \left[ \left( 3 \sin x - 2 \right)^2 - 4 \right]\]
\[ = 4 - \left( 3 \sin x - 2 \right)^2 \]
\[\text{ Minimum value of } \left( 3 \sin x - 2 \right)^2 \text{ is } 0 . \]
\[\text{ Therefore, maximum value of f }\left( x \right) = 4 - \left( 3 \sin x - 2 \right)^2 \text{ is } 4 . \]
\[\text{ We are given that } 12\sin x - 9 \sin^2 x \text{ will attain its maximum value at } x = \alpha . \]
\[ \therefore 12\sin\alpha - 9 \sin^2 \alpha = 4\]
\[ \Rightarrow - 9 \sin^2 \alpha + 12\sin\alpha - 4 = 0\]
\[ \Rightarrow 9 \sin^2 \alpha - 12 \sin\alpha + 4 = 0\]
\[ \Rightarrow 9 \sin^2 \alpha - 6\sin\alpha - 6\sin\alpha + 4 = 0\]
\[ \Rightarrow 3\sin\alpha\left( 3\sin\alpha - 2 \right) - 2\left( 3\sin\alpha - 2 \right) = 0\]
\[ \Rightarrow \left( 3\sin\alpha - 2 \right)\left( 3\sin\alpha - 2 \right) = 0\]
\[ \therefore \sin\alpha = \frac{2}{3}\]
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