Find the value of x in each of the following :

Solution:

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,

$x=15^{\circ}$