Question

$If6{\sin }^2\theta -11\sin \theta +4=0,then\sin \theta$ is $\frac{k}{2}$, find k

Answer 1

$6{\sin }^2\theta -11\sin \theta +4=0$

$\sin \theta =\frac{11\pm \sqrt{{11}^2-4\cdot 6\cdot 4}}{2\times 6}$

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

$=\frac{11\pm \sqrt{25}}{12}$

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

$Sin\theta =\frac{11+5}{12}andSin\theta =\frac{11-5}{12}$

$Sin\theta =\frac{16}{12}=\frac{4}{3}andSin\theta =\frac{6}{12}=\frac{1}{2}$

$Sin\theta =\frac{4}{3}\left(NotPossible\right)$

$sin\theta =\frac{1}{2}$

Comparing with $\frac{k}{2}$

$\therefore k=1$