Question

If $${\cos }^{-1}\left(\cos x\right)=x$$ is satisfied by
A.$$x\in R$$
B.$$x\in \left[0,\pi \right]$$
C.$$x\in \left[-1,1\right]$$
D.none of these

Answer 1

Hint: The inverse trigonometric functions are also known as Arc functions. Inverse Trigonometric Functions are defined in a certain interval under restricted domains. Trigonometry basics include the basic trigonometry and trigonometric ratios such as sin, cos, tan, cosec, sec, cot and we know that the domain of $${\cos }^{-1}x$$ is $$\left[-1,1\right]$$ and hence with respect to $$\cos x$$ we can find the range of $$x$$ .

Complete step by step solution:
Given,
 $${\cos }^{-1}\left(\cos x\right)=x$$
We know that the domain of $${\cos }^{-1}x$$ is $$\left[-1,1\right]$$
 $$\Rightarrow$$ $$-1⩽\cos x⩽1$$
 $$\Rightarrow$$ $${\cos }^{-1}\left(1\right)⩽{\cos }^{-1}\left(\cos x\right)⩽{\cos }^{-1}\left(-1\right)$$
 $$\Rightarrow$$ $$0⩽{\cos }^{-1}\left(\cos x\right)⩽\pi$$
If, $$0⩽x⩽\pi$$ then
 $$x\in \left[0,\pi \right]$$ .
Therefore, option B is the right answer.
So, the correct answer is “Option B”.

Note: The key point to evaluate any inverse trigonometric function is that we must know all the basic trigonometric functions and their inverse relationship. As in the given equation consists of cosine function and inverse function of cos, i.e., $${\displaystyle \frac{d}{dx}}\arccos \left(x\right)=-{\displaystyle \frac{1}{\sqrt{1-x^2}}}$$ ,hence we must know all the inverse trigonometric identities with respect to the range of function to solve these types of questions.