Question

If $12sin\theta -9si{n}^2\theta$ attains its maximum value of $\theta =\alpha ,$ then write the value of $sin\alpha$.

Answer 1

$12sin\theta -9si{n}^2\theta$ has maximum value as 4, so
$12sin\alpha -9si{n}^2\alpha =4$
$9si{n}^2\alpha -12sin\alpha +4=0$
$9si{n}^2\alpha -6sin\alpha -6sin\alpha +4=0$
$3sin\alpha (3sin\alpha -2)-2(3sin\alpha -2)=0$
$(3sin\alpha -2)(3sin\alpha -2)=0$
$3sin\alpha -2=0$
$sin\alpha =\frac{2}{3}$