Question

Modulus function : Let $f:R\to R$ be given by $f\left(x\right)=|x|$ for each $x\in R$, then

$f\left(x\right)=|x|=\{\begin{array}{cc}x& ,x\ge 0\\ -x& ,x<0\end{array}$
The number of solution of the equation $|\cos x-\sin x|=2\cos x$ in $[0,2\pi ]$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

$|\cos x-\sin x|=2\cos x$ ... (1)

$⟹\cos x-\sin x=\pm 2\cos x$

$⟹\cos x-\sin x=2\cos x\text{and/or}\cos x-\sin x=-2\cos x$

$⟹-\sin x=\cos x\text{and/or}-\sin x=-3\cos x$

$⟹\tan x=-1\text{and/or}\tan x=3$

$⟹x=\frac{3\pi }{4}\text{or}\frac{7\pi }{4}\text{and/or}x={\tan }^{-1}3$

From definition of modulus function and equation (1), we know that $2\cos x\ge 0⟹x\in [0,\frac{\pi }{2}]\cup [\frac{3\pi }{2},2\pi ]$

Hence, only feasible solution exists when $x=\frac{7\pi }{4}\text{or}x={\tan }^{-1}\in [0,\frac{\pi }{2}]$

Hence, 2 solutions exist.