Question

Find the acute angle $\theta$(principal value in degrees) such that :
$2\sin \theta -\sqrt{3}\sec \theta +1=0$

Answer 1

$2\sin \theta -\sqrt{3}\sec \theta +1=0$

$\Rightarrow (2\sin \theta +1{)}^2=\frac{3}{{\cos }^2\theta }$

$\Rightarrow (1-{\sin }^2\theta )(4{\sin }^2\theta +4\sin \theta +1)=3$

$\Rightarrow 4{\sin }^2\theta +4\sin \theta +1-4{\sin }^4\theta -4{\sin }^3\theta -{\sin }^2\theta =3$

$\Rightarrow 4{\sin }^4\theta +4{\sin }^3\theta -4\sin \theta +2=0$

$\Rightarrow {\sin }^2\theta (4{\sin }^2\theta +4\sin \theta -3)-2(2\sin \theta -1)=0$

$\Rightarrow {\sin }^2\theta (4{\sin }^2\theta +4\sin \theta -3)-2(2\sin \theta -1)=0$
$\Rightarrow {\sin }^2\theta (2\sin \theta -1)(2\sin \theta -3)-2(2\sin \theta -1)=0$

$\Rightarrow (2\sin \theta -1)(2{\sin }^3\theta -3{\sin }^2\theta -2)=0$

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

$\theta ={30}^{\circ }=\frac{\pi }{6}$