Question

If$x \ne \dfrac{{n\pi }}{2}$ and ${\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = 1$ then all solutions of $x$ are given by
A. $2n\pi + \dfrac{\pi }{2}$
B. $\left( {2n + 1} \right)\pi - \dfrac{\pi }{2}$
C. $n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}$
D. None of these

Answer 1

First, we shall analyze the given information so that we are able to solve the problem. Here we are given that $x \ne \dfrac{{n\pi }}{2}$ and${\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = 1$. And, we are asked to calculate the solution of $x$. We all know that anything with a power of zero is one. So we need to apply${1^0} = {\left( {\cos x} \right)^0}$in the given equation. Then we need to analyze the obtained solution whether it will be the correct solution or not.

Complete step by step answer:
It is given that${\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = 1$
It is given that$x \ne \dfrac{{n\pi }}{2}$
When$n = 1$ $x \ne \dfrac{\pi }{2}$ which implies$\cos x \ne \cos \dfrac{\pi }{2}$
Hence we get$\cos x \ne 0$ .
When$n = 0$ $x \ne 0$ which implies$\cos x \ne \cos 0$
Hence we get$\cos x \ne 1$
We are asked to calculate the solution of $x$
${\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = 1 \Rightarrow {\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = {\left( {\cos x} \right)^0}$ (Here we applied${1^0} = {\left( {\cos x} \right)^0}$
$\Rightarrow {\left( {\cos x} \right)^{{{\sin }^2}x - 3\sin x + 2}} = {\left( {\cos x} \right)^0}$
Since the bases are equal, we shall compare the exponents/power of$\cos x$
$\Rightarrow {\sin ^2}x - 3\sin x + 2 = 0$
Now, we need to split the middle term.
$\Rightarrow {\sin ^2}x - \sin x - 2\sin x + 2 = 0$
We shall pick the common terms.
$\Rightarrow \sin x\left( {\sin x - 1} \right) - 2\left( {\sin x - 1} \right) = 0$
$\Rightarrow \left( {\sin x - 1} \right)\left( {\sin x - 2} \right) = 0$
Hence, $\left( {\sin x - 1} \right) = 0or\left( {\sin x - 2} \right) = 0$
That is $\sin x = 1or\sin x = 2$
Here $\sin x = 2$is not possible.
Now, consider$\sin x = 1$
$\sin x = 1 \Rightarrow \cos x = 0$
It is given that$x \ne \dfrac{{n\pi }}{2}$
When$n = 1$ $x \ne \dfrac{\pi }{2}$ which implies$\cos x \ne \cos \dfrac{\pi }{2}$
Hence we get$\cos x \ne 0$
Hence $\sin x = 1$is also not possible.
Hence there is no solution for $x$.

So, the correct answer is “Option D”.

Note: Here we said that$\sin x = 2$is not possible. The sine wave exists between minus one and one (i.e. $- 1 \leqslant \sin x \leqslant 1$). Since the sine lies between minus one and one, the value of sine two is not applicable. Hence, we neglect the solution$\sin x = 2$.
Also, we are given that$x \ne \dfrac{{n\pi }}{2}$. We shall substitute$n = 1$, we obtain$x \ne \dfrac{\pi }{2}$. This implies $\cos x \ne \cos \dfrac{\pi }{2}$ . That is$\cos x \ne 0$ . Here $\cos x \ne 0$implies$\sin x \ne 1$. So we neglect this solution too. Hence we get no solution for$x$.
Here option 1) is$2n\pi + \dfrac{\pi }{2}$. When$n = 1$, $\sin 2\pi + \dfrac{\pi }{2} = \sin \dfrac{{5\pi }}{2} = 1$
Hence we cannot choose the option$2n\pi + \dfrac{\pi }{2}$.
Here option 2) is$\left( {2n + 1} \right)\pi - \dfrac{\pi }{2}$. When$n = 1$, $\sin \left( {3n} \right)\pi - \dfrac{\pi }{2} = \sin \dfrac{{5\pi }}{2} = 1$
Hence we cannot choose the option$\left( {2n + 1} \right)\pi - \dfrac{\pi }{2}$.
Here option 3) is$n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}$. When$n = 1$, $\sin \pi - \dfrac{\pi }{2} = \sin \dfrac{\pi }{2} = 1$
Hence we cannot choose the option$n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}$.
Therefore, we chose none of these.