Question

# Find the value of$x$ in the following:$\sqrt 3 \tan 2x = \cos {60^0} + \sin {45^0}\cos {45^0}$

Given $\sqrt 3 \tan 2x = \cos {60^0} + \sin {45^0}\cos {45^0}$
$\Rightarrow \sqrt 3 \tan 2x = \dfrac{1}{2} \times \left( {\dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}} \right)$ $\left[ {\because \cos {{60}^0} = \dfrac{1}{2},\cos {{45}^0} = \sin {{45}^0} = \dfrac{1}{{\sqrt 2 }}} \right]$
$\Rightarrow \sqrt 3 \tan 2x = \dfrac{1}{2} + \dfrac{1}{2} \\ \Rightarrow \sqrt 3 \tan 2x = 1 \\ \Rightarrow \tan 2x = \dfrac{1}{{\sqrt 3 }} \\$
${\text{tan3}}{{\text{0}}^0}{\text{ = }}\dfrac{1}{{\sqrt 3 }}$
$\Rightarrow \tan 2x = {\text{tan3}}{{\text{0}}^0} \\ \Rightarrow 2x = {30^0} \\$
$\therefore x = {15^0}$ Answer.