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# Solve the given equation: If ${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$ ,then the value of x is equal toA. 0,$\frac{1}{2}$ B. 1, $\frac{1}{2}$C. 0D.$\frac{1}{2}$ Verified
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Hint-Make use of the formula $\sin \left( {\frac{\pi }{2} + \theta } \right) = \cos \theta$ and solve the problem
The given equation is ${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$,
${\text{On shifting 2}}{\sin ^{ - 1}}x{\text{ to the RHS we get}}$
${\sin ^{ - 1}}(1 - x) = \frac{\pi }{2} + 2{\sin ^{ - 1}}x$
${\text{Further on shifting }}{\sin ^{ - 1}}{\text{ to the RHS, we get}}$
$\left( {1 - x} \right) = \sin \left( {\frac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)$
We know from the formula that $\sin \left( {\frac{\pi }{2} + x} \right) = \cos x$
$So,{\text{ }}we{\text{ }}can{\text{ }}write{\text{ }}the{\text{ }}equation{\text{ }}as\;\;\;the{\text{ }}equation{\text{ }}as\;\;\;(1 - x) = \cos (2{\sin ^{ - 1}}x)$ $But{\text{ we know the result which says 2}}{\sin ^{ - 1}}x = {\cos ^{ - 1}}(1 - 2{x^2})$
$\begin{gathered} {\text{So, from this we get the equation to be }} \\ \Rightarrow {\text{(1 - x) = cos[}}{\cos ^{ - 1}}(1 - 2{x^2})] \\ \end{gathered}$
$We{\text{ know another formula which says cos(}}{\cos ^{ - 1}}x) = x$
$\begin{gathered} {\text{Using this result , we can now write the equation as }} \\ \Rightarrow {\text{(1 - x) = 1 - 2}}{{\text{x}}^2} \\ \Rightarrow 2{x^2} - x = 0 \\ \Rightarrow x(2x - 1) = 0 \\ \Rightarrow x = 0{\text{ or }}x = \frac{1}{{2,}} \\ If{\text{ x = }}\frac{1}{2}, \\ LHS = {\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{2} - 2{\sin ^{ - 1}}\frac{1}{2} = - \frac{\pi }{6} \\ But{\text{ RHS = }}\frac{\pi }{2} \\ LHS \ne RHS \\ So,x = \frac{1}{2}{\text{ is not a solution to the given equation}} \\ \end{gathered}$
$\begin{gathered} If{\text{ x = 0,}} \\ We{\text{ get LHS = }}{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = {\sin ^{ - 1}}1 - 2{\sin ^{ - 1}}0 = \frac{\pi }{2} - 0 = \frac{\pi }{2} \\ \end{gathered}$$So,we{\text{ have LHS = RHS = }}\frac{\pi }{2}$
So, we can write x=0 is the solution for this given equation
So, option C is the correct answer
Note:Whenever we are solving these kinds of problems, we have to always choose the value
of x such that LHS=RHS. If we get any value of x such that we won't get LHS=RHS, then such
value of x should not be considered.

Last updated date: 23rd Sep 2023
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