Question

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

Answer 1

Given $\sqrt{3}\tan 2x=\cos {60}^{\circ }+\sin {45}^{\circ }\cos {45}^{\circ }$

$\sqrt{3}\tan 2x=\frac{1}{2}+(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})$ $[\because \cos {60}^{\circ }=\frac{1}{2},\cos {45}^{\circ }=\sin {45}^{\circ }=\frac{1}{\sqrt{2}}]$

$\sqrt{3}\tan 2x=\frac{1}{2}+\frac{1}{2}$

$\sqrt{3}\tan 2x=1$

$\tan 2x=\frac{1}{\sqrt{3}}$

we know that

$\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

$\tan 2x=\tan {30}^{\circ }$

$2x={30}^{\circ }$

$\therefore x={15}^{\circ }$