# Solve the given equation: If ${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$

,then the value of x is equal to

A. 0,$\frac{1}{2}$

B. 1, $\frac{1}{2}$

C. 0

D.$\frac{1}{2}$

Last updated date: 30th Mar 2023

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Answer

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309.9k+ views

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.

\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.

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