# Domain of $f\left( x \right) = y = \sqrt {{{\log }_3}\left[ {\cos \left( {\sin x} \right)} \right]} $ is

\[

{\text{A}}{\text{. }}3\left\{ {\dfrac{{2\pi }}{2}:n \in Z} \right\} \\

{\text{B}}{\text{. }}\left\{ {2n\pi :n \in Z} \right\} \\

{\text{C}}{\text{. }}\left\{ {n\pi :n \in Z} \right\} \\

\]

${\text{D}}{\text{. }}$ None of these

Last updated date: 17th Mar 2023

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Answer

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Hint- Here, we will find the values of $x$ corresponding to which the given function will be defined.

The given function is $f\left( x \right) = y = \sqrt {{{\log }_3}\left[ {\cos \left( {\sin x} \right)} \right]} $

We have to find the domain of the above given function.

As we know that the value of any function inside the square root should always be greater than or equal to zero else the function will not be defined.

i.e., ${\log _3}\left[ {\cos \left( {\sin x} \right)} \right] \geqslant 0{\text{ }} \to {\text{(1)}}$

Now, solve the above inequality for the values of $x$

Taking antilog of the inequality (1), we have

$ \Rightarrow \left[ {\cos \left( {\sin x} \right)} \right] \geqslant {3^0} \Rightarrow \cos \left( {\sin x} \right) \geqslant 1$

Also, we know that the value of cosine of any angle $\theta $ lies between $ - 1$ and $1$

i.e., $ - 1 \leqslant \cos \theta \leqslant 1$

$\cos \left( {\sin x} \right) = 1 \Rightarrow \sin x = 0 \Rightarrow x = n\pi $, where $n \in Z$.

So, domain of the given function is \[\left\{ {n\pi :n \in Z} \right\}\]

Therefore, option C is correct.

Note- Domain of any function of variable $x$ are the values of $x$ for which the function will be defined. In this particular problem, we have considered only $\sin x = 0$ when $\cos \left( {\sin x} \right) = 1$ because other values will also correspond to the same result.

The given function is $f\left( x \right) = y = \sqrt {{{\log }_3}\left[ {\cos \left( {\sin x} \right)} \right]} $

We have to find the domain of the above given function.

As we know that the value of any function inside the square root should always be greater than or equal to zero else the function will not be defined.

i.e., ${\log _3}\left[ {\cos \left( {\sin x} \right)} \right] \geqslant 0{\text{ }} \to {\text{(1)}}$

Now, solve the above inequality for the values of $x$

Taking antilog of the inequality (1), we have

$ \Rightarrow \left[ {\cos \left( {\sin x} \right)} \right] \geqslant {3^0} \Rightarrow \cos \left( {\sin x} \right) \geqslant 1$

Also, we know that the value of cosine of any angle $\theta $ lies between $ - 1$ and $1$

i.e., $ - 1 \leqslant \cos \theta \leqslant 1$

$\cos \left( {\sin x} \right) = 1 \Rightarrow \sin x = 0 \Rightarrow x = n\pi $, where $n \in Z$.

So, domain of the given function is \[\left\{ {n\pi :n \in Z} \right\}\]

Therefore, option C is correct.

Note- Domain of any function of variable $x$ are the values of $x$ for which the function will be defined. In this particular problem, we have considered only $\sin x = 0$ when $\cos \left( {\sin x} \right) = 1$ because other values will also correspond to the same result.

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