Question

If $x = \dfrac{{n\pi }}{2}$ satisfies the equation $\sin \dfrac{x}{2} - \cos \dfrac{x}{2} = 1 - \sin x$ and the inequality $\left| {\dfrac{x}{2} - \dfrac{\pi }{2}} \right| \le \dfrac{{3\pi }}{4}$, then
A) $n = - 1,0,3,5$
B) $n = 1,2,4,5$
C) $n = 0,2,4$
D) $n = - 1,1,3,5$

Answer 1

Here we will first find the values of $x$ by solving the given inequality. Then we will substitute the values of $x$ in the given equation. As we substitute those values, we will find the required values of $n$.

Complete step by step solution:
At first we have to solve the inequality $\left| {\dfrac{x}{2} - \dfrac{\pi }{2}} \right| \le \dfrac{{3\pi }}{4}$ to find the value of $x$.
Here, modulus represents the non-negative integer without regard to its sign.
$\Rightarrow \left| {\dfrac{x}{2} - \dfrac{\pi }{2}} \right| \le \dfrac{{3\pi }}{4}$
Taking modulus on both the sides, we get
$\Rightarrow \dfrac{x}{2} - \dfrac{\pi }{2} \le \left| {\dfrac{{3\pi }}{4}} \right|$
$\Rightarrow \dfrac{{x - \pi }}{2} \le \dfrac{{3\pi }}{4}$
Multiplying both side by 2, we get
$\Rightarrow x - \pi \le \dfrac{{3\pi }}{2}$
Adding $\pi$ on both the sides, we get
$\Rightarrow x \le \dfrac{{3\pi }}{2} + \pi$
$\Rightarrow x \le \dfrac{{5\pi }}{2}$
So, the value of $x$ should be less than or equal to $\dfrac{{5\pi }}{2}$ . Since the value lies in the modulus value, the negative value of $n$ can be neglected. So the values of $x$ are $x = 0,\dfrac{\pi }{2},\dfrac{{2\pi }}{2},\dfrac{{3\pi }}{2},\dfrac{{4\pi }}{2},\dfrac{{5\pi }}{2}$ that is $x = 0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi ,\dfrac{{5\pi }}{2}$ .These values of $x$ are obtained by the equation $x = \dfrac{{n\pi }}{2}$.
These values of $x$ are substituted in the equation $\sin \dfrac{x}{2} - \cos \dfrac{x}{2} = 1 - \sin x$ to obtain the values of $n$.
Substituting $x = 0$ in the given equation, we get
$\sin 0 - \cos 0 = 1 - \sin 0$
Substituting the values of $\sin 0$ and $\cos 0$, we get
$\begin{array}{l} \Rightarrow 0 - 1 = 1 - 0\\\ \Rightarrow - 1 = 1\end{array}$
$x = 0$ does not satisfy the equation.
Substituting $x = \dfrac{\pi }{2}$ in the given equation, we get
$\sin \dfrac{\pi }{4} - \cos \dfrac{\pi }{4} = 1 - \sin \dfrac{\pi }{2}$
Substituting the values of $\sin \dfrac{\pi }{2}$ and $\cos \dfrac{\pi }{2}$, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }} = 1 - 1\\\ \Rightarrow 0 = 0\end{array}$
$x = \dfrac{\pi }{2}$ satisfies the equation.
Substituting $x = \pi$ in the given equation, we get
$\sin \dfrac{\pi }{2} - \cos \dfrac{\pi }{2} = 1 - \sin \pi$
Substituting the values of $\sin \pi$ and $\cos \pi$, we get
$\begin{array}{l} \Rightarrow 1 - 0 = 1 - 0\\\ \Rightarrow 1 = 1\end{array}$
$x = \pi$ satisfies the equation.
Substitute $x = \dfrac{{3\pi }}{2}$ in the given equation, we get
$\sin \dfrac{{3\pi }}{4} - \cos \dfrac{{3\pi }}{4} = 1 - \sin \dfrac{{3\pi }}{2}$
Substituting the values of $\sin \dfrac{{3\pi }}{2}$ and $\cos \dfrac{{3\pi }}{2}$, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} = 1 + 1\\\ \Rightarrow \sqrt 2 = 2\end{array}$
$x = \dfrac{{3\pi }}{2}$does not satisfy the equation.
Substituting $x = 2\pi$ in the given equation, we get
$\sin \pi - \cos \pi = 1 - \sin 2\pi$
Substituting the values of $\sin 2\pi$ and $\cos 2\pi$, we get
$\begin{array}{l} \Rightarrow 0 + 1 = 1 - 0\\\ \Rightarrow 1 = 1\end{array}$
$x = 2\pi$ satisfies the equation.
Substitute $x = \dfrac{{5\pi }}{2}$ in the given equation, we get
$\sin \dfrac{{5\pi }}{4} - \cos \dfrac{{5\pi }}{4} = 1 - \sin \dfrac{{5\pi }}{2}$
Substituting the values of $\sin \dfrac{{3\pi }}{2}$ and $\cos \dfrac{{3\pi }}{2}$, we get
$\begin{array}{l} \Rightarrow \dfrac{{ - 1}}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} = 1 - 1\\\ \Rightarrow 0 = 0\end{array}$
$x = \dfrac{{5\pi }}{2}$ satisfies the equation.
So the values of $x = \dfrac{\pi }{2},\pi ,2\pi ,\dfrac{{5\pi }}{2}$ .
By comparing the coefficient of $x$ , we get $n = 1,2,4,5$

Therefore, $n = 1,2,4,5$

Note:
Note: Here, we need to keep in mind different values of the trigonometric function of $\sin \theta$ and $\cos \theta$. If we do not remember the values we might substitute wrong values in the equation and hence get wrong answers. We might commit a mistake by leaving the solution after solving the inequality and getting the value of $x$. This will give us the wrong values of $n$. We need to find all values of $x$ which satisfies the given equation to find the correct values of $n$.