If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
Let $f(x)=12 \sin x-9 \sin ^{2} x$
$=-\left(9 \sin ^{2} x-12 \sin x\right)$
$=-\left[(3 \sin x)^{2}-2.3 \sin x .2+2^{2}-4\right]$
$=-\left[(3 \sin x-2)^{2}-4\right]$
$=4-(3 \sin x-2)^{2}$
Minimum value of $(3 \sin x-2)^{2}$ is 0 .
Therefore, maximum value of $f(x)=4-(3 \sin x-2)^{2}$ is 4 .
We are given that $12 \sin x-9 \sin ^{2} x$ will attain its maximum value at $x=\alpha$.
$\therefore 12 \sin \alpha-9 \sin ^{2} \alpha=4$
$\Rightarrow-9 \sin ^{2} \alpha+12 \sin \alpha-4=0$
$\Rightarrow 9 \sin ^{2} \alpha-12 \sin \alpha+4=0$
$\Rightarrow 9 \sin ^{2} \alpha-6 \sin \alpha-6 \sin \alpha+4=0$
$\Rightarrow 3 \sin \alpha(3 \sin \alpha-2)-2(3 \sin \alpha-2)=0$
$\Rightarrow(3 \sin \alpha-2)(3 \sin \alpha-2)=0$
$\therefore \sin \alpha=\frac{2}{3}$
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