Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{4\pi }{3}$
• D. $\frac{5\pi }{3}$

Answer 1

The correct option is C $\frac{4\pi }{3}$

Given equation is

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

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

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

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

Let $\sin \theta =x$

$⟹2x^2-\sqrt{3}x-3=0$

$2x^2-2\sqrt{3}x+\sqrt{3}x-3=0$

$2x(x-\sqrt{3})+\sqrt{3}(x-\sqrt{3})=0$

$(2x+\sqrt{3})(x-\sqrt{3})=0$

$⟹x=\sqrt{3},-\frac{\sqrt{3}}{2}$

$⟹\sin \theta =-\frac{\sqrt{3}}{2}$ it cannot be $\sqrt{3}$

$⟹\sin \theta =\sin \frac{4\pi }{3}$

$⟹\theta =\frac{4\pi }{3}$