Question

If $6{\sin }^2\theta -11\sin \theta +4=0$, then find $\sin \theta$.

Answer 1

Let by the quadratic formula :

$sin\theta =\frac{11\pm \sqrt{121-96}}{12}$

$=\frac{11\pm 5}{12}$

Implies that,

$sin\theta =\frac{4}{3},\frac{1}{3}$

therefore,

$\theta =si{n}^{-1}\frac{4}{3},si{n}^{-1}\frac{1}{2}$