Question

The general solution of $\left| \sin x \right|=\cos x$ is (When $n\in Z $) given by
(A) $n\pi +\dfrac{\pi }{4}$
(B) $2n\pi \pm \dfrac{\pi }{4}$
(C) $n\pi \pm \dfrac{\pi }{4}$
(D) $n\pi -\dfrac{\pi }{4}$

Answer 1

Hint: We solve this question by first considering the given equation, $\left| \sin x \right|=\cos x$. Then we consider the definition of modulus function and then we use it to define $\left| \sin x \right|$. Then we consider the case when $\sin x>0$. Then we solve it to find the value of $\tan x$. Then we use the formula for the general solution of equation $\tan x=\tan \theta $, $x=n\pi +\theta $ and find the general solution when $\sin x>0$. Then we consider the case when $\sin x<0$ and solve it to find the value of $\tan x$. Then we use the formula for the general solution of equation $\tan x=\tan \theta $, $x=n\pi +\theta $ and find the general solution when $\sin x<0$. So, combining the obtained solutions we can find the general solution of $\left| \sin x \right|=\cos x$.

Complete step by step answer:
The equation we are given is $\left| \sin x \right|=\cos x$.
First let us consider the definition of modulus function.
$\left| x \right|=\left\{ \begin{matrix}
   x\ \ \ if\ \ x>0 \\
   -x\ \ \ if\ \ x<0 \\
\end{matrix} \right.$
So, we can write $\left| \sin x \right|$ as,
$\left| \sin x \right|=\left\{ \begin{matrix}
   \sin x\ \ \ if\ \ \sin x>0 \\
   -\sin x\ \ \ if\ \ \sin x<0 \\
\end{matrix} \right...............\left( 1 \right)$
First let us consider the case when $\sin x>0$.
Then from equation (1) we get $\left| \sin x \right|$ as,
$\Rightarrow \left| \sin x \right|=\sin x$
As we are given that $\left| \sin x \right|=\cos x$, we get,
$\begin{align}
  & \Rightarrow \cos x=\sin x \\
 & \Rightarrow \dfrac{\sin x}{\cos x}=1 \\
 & \Rightarrow \tan x=1 \\
\end{align}$
Now let us find the principal value of the above tangent function.
$\Rightarrow \tan x=\tan \dfrac{\pi }{4}$
Now let us consider the formula for the general solution of equation $\tan x=\tan \theta $,
$x=n\pi +\theta $
So, we get the general solution of $\tan x=\tan \dfrac{\pi }{4}$ as,
$\Rightarrow x=n\pi +\dfrac{\pi }{4}...........\left( 2 \right)$
Now let us consider the case when $\sin x<0$.
Then from equation (1) we get $\left| \sin x \right|$ as,
$\Rightarrow \left| \sin x \right|=-\sin x$
As we are given that $\left| \sin x \right|=\cos x$, we get,
$\begin{align}
  & \Rightarrow \cos x=-\sin x \\
 & \Rightarrow \dfrac{\sin x}{\cos x}=-1 \\
 & \Rightarrow \tan x=-1 \\
\end{align}$
Now let us find the principal value of the above tangent function.
$\Rightarrow \tan x=\tan \left( -\dfrac{\pi }{4} \right)$
Now let us consider the formula for the general solution of equation $\tan x=\tan \theta $,
$x=n\pi +\theta $
So, we get the general solution of $\tan x=\tan \left( -\dfrac{\pi }{4} \right)$ as,
$\Rightarrow x=n\pi -\dfrac{\pi }{4}...........\left( 2 \right)$
So, from equation (2) and equation (3) we get the general solution of $\left| \sin x \right|=\cos x$ as,
$\begin{align}
  & \Rightarrow x=n\pi +\dfrac{\pi }{4},n\pi -\dfrac{\pi }{4} \\
 & \Rightarrow x=n\pi \pm \dfrac{\pi }{4} \\
\end{align}$
So, we get the general solution of $\left| \sin x \right|=\cos x$ as $x=n\pi \pm \dfrac{\pi }{4}$.

So, the correct answer is “Option A”.

Note: The common mistake one makes is one might take the formula for the general solution of equation $\tan x=\tan \theta $ as $x=2n\pi +\theta $ then one gets the final solution for the question as $x=2n\pi \pm \dfrac{\pi }{4}$ and mark the answer as Option B. But it is wrong as the correct formula is $x=n\pi +\theta $.