Question

Solve: $1+{\sin }^2\theta =3\sin \theta \cos \theta$

Answer 1

$1+{\sin }^2\theta =3\sin \theta \cos \theta$

$1+{\sin }^2\theta -3\sin \theta \cos \theta =0$

${\sin }^2\theta +{\cos }^2\theta +{\sin }^2\theta -3\sin \theta \cos \theta =0$

since, ${\sin }^2\theta +{\cos }^2\theta =1$

$2{\sin }^2\theta -2\sin \theta \cos \theta +{\cos }^2\theta -\sin \theta \cos \theta =0$

$2\sin \theta (\sin \theta -\cos \theta )-\cos \theta (\sin \theta -\cos \theta )=0$

$(\sin \theta -\cos \theta )(2\sin \theta -\cos \theta )=0$

therefore,

$(\sin \theta -\cos \theta )=0$ .......(i)

or

$(2\sin \theta -\cos \theta )=0$ .......(ii)

from eqn. (i)

$(\sin \theta -\cos \theta )=0$

$\tan \theta =1$

$\theta =\frac{\pi }{4}$

from eqn. (ii)

$(2\sin \theta -\cos \theta )=0$

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

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