Find the value of x in each of the following :
$\sqrt{3} \tan 2 x=\cos 60^{\circ}+\sin 45^{\circ} \cos 45^{\circ}$
We have,
$\sqrt{3} \tan 2 x=\cos 60^{\circ}+\sin 45^{\circ} \cos 45^{\circ} \ldots \ldots$ (1)
Now we know that
$\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$ and $\cos 60^{\circ}=\frac{1}{2}$
Now by substituting above values in equation (1), we get,
$\sqrt{3} \tan 2 x=\cos 60^{\circ}+\sin 45^{\circ} \cos 45^{\circ}$
$\sqrt{3} \tan 2 x=\frac{1}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=\frac{1}{2}+\frac{1}{\sqrt{2} \times \sqrt{2}}$
$=\frac{1}{2}+\frac{1}{2}$
$=\frac{1+1}{2}$
$=\frac{2}{2}$
$=1$
Therefore,
$\sqrt{3} \tan 2 x=1$
$\Rightarrow \tan 2 x=\frac{1}{\sqrt{3}}$.....(2)
Since,
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}$....(3)
Therefore by comparing equation (2) and (3)
We get,
$2 x=30^{\circ}$
$\Rightarrow x=\frac{30^{\circ}}{2}$
$\Rightarrow x=15^{\circ}$
Therefore,
$x=15^{\circ}$