If 12 sin x − 9sin

Solution:

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 $x=\alpha$.

$\therefore 12 \sin \alpha-9 \sin ^{2} \alpha=4$

$\Rightarrow-9 \sin ^{2} \alpha+12 \sin \alpha-4=0$

$\Rightarrow 9 \sin ^{2} \alpha-12 \sin \alpha+4=0$

$\Rightarrow 9 \sin ^{2} \alpha-6 \sin \alpha-6 \sin \alpha+4=0$

$\Rightarrow 3 \sin \alpha(3 \sin \alpha-2)-2(3 \sin \alpha-2)=0$

$\Rightarrow(3 \sin \alpha-2)(3 \sin \alpha-2)=0$

$\therefore \sin \alpha=\frac{2}{3}$