Question

Find the value of x in the following :
$\cos 2x=\cos {60}^{\circ }\cos {30}^{\circ }+\sin {60}^{\circ }\sin {30}^{\circ }$

Answer 1

Given $\cos 2x=\cos {60}^{\circ }\cos {30}^{\circ }+\sin {60}^{\circ }\sin {30}^{\circ }$

$\cos 2x=(\frac{1}{2}\times \frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2}\times \frac{1}{2})$ $[\because \sin {30}^{\circ }=\cos {60}^{\circ }=\frac{1}{2},\cos {30}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}]$

$\cos 2x=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$

$\cos 2x=2\left(\frac{\sqrt{3}}{4}\right)$

$\cos 2x=\frac{\sqrt{3}}{2}$

we know that,

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

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

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

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