Question

Under root 3 sin theta = cos theta then the value of 3 cos square theta + 2 cos theta upon 3 cos theta is

Answer 1

Answer:

$\rm \dfrac{9+4\sqrt{3}}{6\sqrt{3}}$

Step-by-step explanation:

$\rm \sqrt{3} \sin θ = \cos θ$

$\rm \sqrt{3} = \dfrac{\cos θ}{\sin θ}$

$\rm \cot 30^{\circ} = \cot θ$

$\rm θ=30^{\circ}$

$\rm \dfrac{3 \cos ^2 θ + 2 \cos θ}{3 \cos θ}$

$\rm \dfrac{3 \cos ^2 30^{\circ} + 2 \cos 30^{\circ}}{3 \cos 30^{\circ}}$

$\rm \dfrac{3 \bigg( \dfrac{\sqrt{3}}{2} \bigg) ^2+ 2 \bigg( \dfrac{\sqrt{3}}{2} \bigg)}{3 \bigg( \dfrac{\sqrt{3}}{2} \bigg)}$

$\rm \dfrac{ \bigg(\dfrac{9}{4} \bigg) + \sqrt{3}}{ \bigg(\dfrac{3 \sqrt{3}}{2} \bigg)}$

$\rm \dfrac{\bigg( \dfrac{9+4\sqrt{3} }{4}\bigg)}{\bigg( \dfrac{3\sqrt{3}}{2} \bigg)}$

$\rm \dfrac{\bigg( \dfrac{9+4\sqrt{3}}{2}\bigg)}{3\sqrt{3}}$

$\rm \dfrac{9+4\sqrt{3}}{6\sqrt{3}}$