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

Given:

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

To do:

We have to find the value of \( x \).

Solution:  

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

We know that,

$\cos 60^{\circ}=\frac{1}{2}$

$\sin 45^{\circ}=\frac{1}{\sqrt2}$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

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

$\Rightarrow \sqrt{3} \tan 2 x=\frac{1}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt2}$

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

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

$\Rightarrow \tan 2 x=\frac{1}{\sqrt3}$

$\Rightarrow \tan 2 x=\tan 30^{\circ}$          (Since $\tan 30^{\circ}=\frac{1}{\sqrt3}$)

Comparing on both sides, we get,

$2x=30^{\circ}$

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

$x=15^{\circ}$

Hence, the value of $x$ is $15^{\circ}$. 

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Updated on: 10-Oct-2022

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