Question

If we are given $x\ne \dfrac{n\pi }{2}$ and ${{\left| \cos x \right|}^{-{{\sin }^{2}}x-3\sin x+2}}=1$ then all the 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

Hint: To find the value of x for which ${{\left| \cos x \right|}^{-{{\sin }^{2}}x-3\sin x+2}}=1$ , we can first take $\log $ of the whole expression and then two cases will arise that we need to consider:
1) When $\cos x=\pm 1$
2) When $-{{\sin }^{2}}x-3\sin x+2=0$
After that, we have to combine the values of x for which the above two conditions are satisfied and we will get our final answer.

Complete step-by-step solution:
According to question,
${{\left| \cos x \right|}^{-{{\sin }^{2}}x-3\sin x+2}}=1$
Taking $\log $ of the above equation, we get
$\begin{align}
  & \Rightarrow \log \left( {{\left| \cos x \right|}^{-{{\sin }^{2}}x-3\sin x+2}} \right)=\log 1 \\
 & \because \log {{m}^{n}}=n\log m,\log 1=0 \\
 & \Rightarrow \left( -{{\sin }^{2}}x-3\sin x+2 \right)\log \left( \left| \cos x \right| \right)=0 \\
\end{align}$
So, either we can have $-{{\sin }^{2}}x-3\sin x+2=0$ or $\log \left( \left| \cos x \right| \right)=0$
We will first consider case 1 i.e.
$\log \left( \left| \cos x \right| \right)=0$
$\Rightarrow \cos x=\pm 1$
The above equation is satisfied when we put an integral multiple of $\pi $ in place of x.
$\Rightarrow x=n\pi $
Now considering the second case i.e.
$\begin{align}
  & -{{\sin }^{2}}x-3\sin x+2=0 \\
 & \Rightarrow {{\sin }^{2}}x+3\sin x-2=0 \\
\end{align}$
We know that the quadratic formula for equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now, here we have $x=\sin x,a=1,b=3,c=-2$ . Applying quadratic formula to solve the equation, we get
$\begin{align}
  & \Rightarrow \sin x=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\times \left( -2 \right)}}{2} \\
 & \Rightarrow \sin x=\dfrac{-3\pm \sqrt{9+8}}{2} \\
 & \Rightarrow \sin x=\dfrac{-3\pm \sqrt{17}}{2} \\
 & \Rightarrow \sin x=-3.56\text{ or }0.56 \\
\end{align}$
Since the value of $\sin x$ is always between -1 and 1, therefore $\sin x=-3.56$ is rejected.
We know that the general solution of $\sin x=\sin y$ is given as
$x=n\pi +{{\left( -1 \right)}^{n}}y$
Now the value of x which satisfies $\sin x=0.56$ can be given by
$\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}{{\sin }^{-1}}\left( 0.56 \right)$
Here we have taken $\sin y=0.56$ which makes $y={{\sin }^{-1}}\left( 0.56 \right)$. We will use the result $\left\{ {{\sin }^{-1}}\left( 0.56 \right) \right.=0.582$ .
$\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}0.582$
Combining the values of x for both the cases, we get
$\Rightarrow x=n\pi \text{ or }x=n\pi +{{\left( -1 \right)}^{n}}0.582$
Since the answer doesn’t match with any of the option given to us.
Therefore, option (d) is the correct option.

Note: Instead of taking $\log $ , we could have used the fact that ${{1}^{n}}=1$ and ${{a}^{0}}=1$ . By comparing the expression with these we could have saved some of our time in solving the question. Sometimes you should try putting options in the expression in order to get the answer quickly or to verify the answer you got, this way you can be more accurate.