Solveeit Logo
Login

Question

Question: Solve the given equation: If \({\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}\) ,then ...

Solve the given equation: If sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}
,then the value of x is equal to
A. 0,12\frac{1}{2}
B. 1, 12\frac{1}{2}
C. 0
D.12\frac{1}{2}

Explanation

Solution

Hint-Make use of the formula sin⁡(π2+θ)=cos⁡θ\sin \left( {\frac{\pi }{2} + \theta } \right) = \cos \theta and solve the problem
The given equation is sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2},
On shifting 2sin⁡−1x to the RHS we get{\text{On shifting 2}}{\sin ^{ - 1}}x{\text{ to the RHS we get}}
sin⁡−1(1−x)=π2+2sin⁡−1x{\sin ^{ - 1}}(1 - x) = \frac{\pi }{2} + 2{\sin ^{ - 1}}x
Further on shifting sin⁡−1 to the RHS, we get{\text{Further on shifting }}{\sin ^{ - 1}}{\text{ to the RHS, we get}}
(1−x)=sin⁡(π2+2sin⁡−1x)\left( {1 - x} \right) = \sin \left( {\frac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)
We know from the formula that sin⁡(π2+x)=cos⁡x\sin \left( {\frac{\pi }{2} + x} \right) = \cos x
So, we can write the equation as      the equation as      (1−x)=cos⁡(2sin⁡−1x)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 we know the result which says 2sin⁡−1x=cos⁡−1(1−2x2)But{\text{ we know the result which says 2}}{\sin ^{ - 1}}x = {\cos ^{ - 1}}(1 - 2{x^2})
So, from this we get the equation to be  ⇒(1 - x) = cos[cos⁡−1(1−2x2)] \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 know another formula which says cos(cos⁡−1x)=xWe{\text{ know another formula which says cos(}}{\cos ^{ - 1}}x) = x
Using this result , we can now write the equation as  ⇒(1 - x) = 1 - 2x2 ⇒2x2−x=0 ⇒x(2x−1)=0 ⇒x=0 or x=12, If x = 12, LHS=sin⁡−1(1−x)−2sin⁡−1x=sin⁡−112−2sin⁡−112=\-π6 But RHS = π2 LHS≠RHS So,x=12 is not a solution to the given equation \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}
If x = 0, We get LHS = sin⁡−1(1−x)−2sin⁡−1x=sin⁡−11−2sin⁡−10=π2−0=π2 \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 have LHS = RHS = π2So,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.