# Find the value of$x$ in the following:

\[\sqrt 3 \tan 2x = \cos {60^0} + \sin {45^0}\cos {45^0}\]

Answer

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Hint:-Use standard trigonometric angles and compare to find the value of x.

Given \[\sqrt 3 \tan 2x = \cos {60^0} + \sin {45^0}\cos {45^0}\]

$ \Rightarrow \sqrt 3 \tan 2x = \dfrac{1}{2} \times \left( {\dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}} \right)$ $\left[ {\because \cos {{60}^0} = \dfrac{1}{2},\cos {{45}^0} = \sin {{45}^0} = \dfrac{1}{{\sqrt 2 }}} \right]$

$

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

\Rightarrow \sqrt 3 \tan 2x = 1 \\

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

$

We know that

${\text{tan3}}{{\text{0}}^0}{\text{ = }}\dfrac{1}{{\sqrt 3 }}$

$

\Rightarrow \tan 2x = {\text{tan3}}{{\text{0}}^0} \\

\Rightarrow 2x = {30^0} \\

$

$\therefore x = {15^0}$ Answer.

Note: - In any trigonometric question the key concept is to always remember the value of standard trigonometric angles and substitute those values in the question and compare or find the inverse trigonometric angles to find the solution.

Given \[\sqrt 3 \tan 2x = \cos {60^0} + \sin {45^0}\cos {45^0}\]

$ \Rightarrow \sqrt 3 \tan 2x = \dfrac{1}{2} \times \left( {\dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}} \right)$ $\left[ {\because \cos {{60}^0} = \dfrac{1}{2},\cos {{45}^0} = \sin {{45}^0} = \dfrac{1}{{\sqrt 2 }}} \right]$

$

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

\Rightarrow \sqrt 3 \tan 2x = 1 \\

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

$

We know that

${\text{tan3}}{{\text{0}}^0}{\text{ = }}\dfrac{1}{{\sqrt 3 }}$

$

\Rightarrow \tan 2x = {\text{tan3}}{{\text{0}}^0} \\

\Rightarrow 2x = {30^0} \\

$

$\therefore x = {15^0}$ Answer.

Note: - In any trigonometric question the key concept is to always remember the value of standard trigonometric angles and substitute those values in the question and compare or find the inverse trigonometric angles to find the solution.

Last updated date: 27th Sep 2023

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