Question

The equation $ 2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right) $ is satisfied by
A. $ -1\le x\le 1 $
B. $ 0\le x\le 1 $
C. $ x\ge 1 $
D. $ x\le 1 $

Answer 1

Hint:
 We first try to express that the given both equations are the same expressions. We apply the theorem of submultiple angles and get that both represent the same angle. We try to find the general forms of x for which the inverse exists. We find a combined common solution to solve the problem.

Complete step by step answer:
The given equation is $ 2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right) $ . We assume the variable for the value as
 $ 2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\alpha $ .
We apply inverse formula to find that $ 2{{\cos }^{-1}}x=\alpha $ which gives $ x=\cos \left( \dfrac{\alpha }{2} \right) $ .
We also have $ {{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\alpha $ which gives $ \cos \alpha =2{{x}^{2}}-1 $ .
We have the theorem of submultiple angle where we have $ \cos \alpha =2{{\cos }^{2}}\left( \dfrac{\alpha }{2} \right)-1 $ .
We replace those values from the above and get
 $ \begin{align}
  & \Rightarrow \cos \alpha =2{{\cos }^{2}}\left( \dfrac{\alpha }{2} \right)-1 \\
 & \Rightarrow 2{{x}^{2}}-1=2{{x}^{2}}-1 \\
\end{align} $
We found the general solution would be enough for the equation. It only has to satisfy the inverse part.
Now we know the range of $ \cos \alpha $ for any value of $ \alpha $ would be $ -1\le \cos \alpha \le 1 $ .
This means $ -1\le x\le 1 $ and $ -1\le 2{{x}^{2}}-1\le 1 $ .
We now solve the second equation to find the value of x and get
 $ \begin{align}
  & -1\le 2{{x}^{2}}-1\le 1 \\
 & \Rightarrow 0\le 2{{x}^{2}}\le 2 \\
 & \Rightarrow 0\le {{x}^{2}}\le 1 \\
 & \Rightarrow -1\le x\le 1 \\
\end{align} $
The combined solution of the general equations is $ -1\le x\le 1 $ . The correct option is A.

Note:
We need to remember we always need to show the equality of the equations. Sometimes the conditions will be the solution. The solution for x has to solve both the inverse law and the equation at the same time.