Question

If $\sin \theta +3\cos \theta =\sqrt{2}$, then find the value of $\sin 2\theta$.

Answer 1

Dividing both sides by $\cos \theta$

$\tan \theta +3=\sqrt{2}\sec \theta$

Squaring both sides

${(\tan \theta +3)}^2=2{\sec }^2\theta$

${\tan }^2\theta +6\tan \theta +9=2(1+{\tan }^2\theta )$

${\tan }^2\theta -6\tan \theta -7=0$

$\tan \theta =\frac{-(-6)\pm \sqrt{(-6{)}^2-4\ast 1\ast (-7)}}{2}$

$\therefore \tan \theta =7,-1$

We know
$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$

Putting the values of $\tan \theta$

$\sin 2\theta =\frac{2\ast 7}{1+7^2}=\frac{7}{25}$

$\sin 2\theta =\frac{2\ast (-1)}{1+(-1{)}^2}=-1$

$\therefore \sin 2\theta =\frac{7}{25},-1$