Question

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

find $\sin \theta$

Answer 1

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

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

$=\frac{11\pm \sqrt{25}}{2\cdot 6}$

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

$=\frac{11+5}{12},\frac{11-5}{12}$

$=\frac{16}{12},\frac{6}{12}$

$\therefore \sin \theta =\frac{4}{3},\frac{1}{2}$