Question

If $f\left( x \right)=\left[ \cos x\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right) \right]$ where $\left[ . \right]$ denotes the greatest integer function $\le x$. Then solution of the equation $f\left( x \right)=x$ is:
A. $1$.
B. $-1$.
C. $0$.
D. None of these

Answer 1

In this problem we need to find the solution for the given function. In the given function we can observe that we have trigonometric ratios first one is $\cos x\cos \left( x+2 \right)$, second one is ${{\cos }^{2}}\left( x+1 \right)$. We will consider each term individually and use the trigonometric formulas to simplify each term. For the first term we will use the trigonometric formula $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ . for the second term we will use the trigonometric formula $\cos 2x=2{{\cos }^{2}}x-1$. After simplifying each term, we will substitute them in the given function and simplify the equation to get the result.

Formulas Used:
1. $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$.
2. $\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$.
3. $\cos \left( -\theta \right)=\cos \theta$.

Complete Step by Step Procedure;
Given that, $f\left( x \right)=\left[ \cos x\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right) \right]$.
We can observe that there are two terms in the above function. The first is $\cos x\cos \left( x+2 \right)$, second term is ${{\cos }^{2}}\left( x+1 \right)$.
Considering the first term $\cos x\cos \left( x+2 \right)$.
From the trigonometric formula $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$, we can write the above term as
$\begin{aligned} & \Rightarrow \cos x\cos \left( x+2 \right)=\dfrac{1}{2}\left[ \cos \left( x+x+2 \right)+\cos \left( x-x-2 \right) \right] \\\ & \Rightarrow \cos x\cos \left( x+2 \right)=\dfrac{1}{2}\left[ \cos 2\left( x+1 \right)+\cos 2 \right] \\\ \end{aligned}$
Considering the second term ${{\cos }^{2}}\left( x+1 \right)$.
From the trigonometric formula $\cos 2x=2{{\cos }^{2}}x-1$, we can write the above term as
$\Rightarrow {{\cos }^{2}}\left( x+1 \right)=\dfrac{\cos 2\left( x+1 \right)+1}{2}$
Substituting the both the terms in the given function, then we will get
$\begin{aligned} & \Rightarrow f\left( x \right)=\left[ \dfrac{1}{2}\left( \cos 2\left( x+1 \right)+\cos 2 \right)-\left( \dfrac{\cos 2\left( x+1 \right)+1}{2} \right) \right] \\\ & \Rightarrow f\left( x \right)=\left[ \dfrac{1}{2}\cos 2\left( x+1 \right)+\dfrac{1}{2}\cos 2-\dfrac{1}{2}\cos 2\left( x+1 \right)-\dfrac{1}{2} \right] \\\ & \Rightarrow f\left( x \right)=\left[ \dfrac{\cos 2-1}{2} \right] \\\ \end{aligned}$

We have the trigonometric formula $\cos 2x=1-2{{\sin }^{2}}x$, we can write the above equation as
$\Rightarrow f\left( x \right)=\left[ -{{\sin }^{2}}1 \right]=1$
Hence the above given function has many solutions.

Note:
In this problem we have the multiplication of $\cos$, so we have used the above-mentioned formula. We also have the formula $\cos A\cos B=\dfrac{1}{2}\left[ \sin \left( A+B \right)-\sin \left( A-B \right) \right]$. But in this problem, we don’t use the above formula because for simplification the above equation is not suitable for simplification.