Question

The smallest positive value of $\theta$ which satisfies the equation $2co{s}^2\theta +\sqrt{3}sin\theta +1=0$

Answer 1

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

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

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

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

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

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

$\Rightarrow 2\sin \theta (\sin \theta -\sqrt{3})+\sqrt{3}(\sin \theta -\sqrt{3})=0$

$\Rightarrow (\sin \theta -\sqrt{3})(2\sin \theta +\sqrt{3})=0$

$\Rightarrow \sin \theta =\sqrt{3},-\frac{\sqrt{3}}{2}$

$\Rightarrow \sin \theta =-\frac{\sqrt{3}}{2},\sin \theta \ne \sqrt{3}$ since range of $\sin \theta \in [-1,1]$

$\Rightarrow \theta =-\frac{\pi }{3},\pi +\frac{\pi }{3},2\pi -\frac{\pi }{3}$

$\therefore \theta =-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$

The smallest positive value of $\theta$ is $\frac{4\pi }{3}$