Question

If \[{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\], then \[x\] equals.

Answer 1

Hint:An equation is given, and we are asked to find the value of \[x\] for this equation. First try to simplify the given equation further so that you can easily find the value of \[x\]. For simplification, use trigonometric identities. After getting the value of \[x\], check whether it satisfies the given equation.

Complete step by step solution:
Given, the equation \[{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\] (i)
To find the value of \[x\] let us simplify the equation further,
\[{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\]
\[ \Rightarrow {\sin ^{ - 1}}\left( {1 - x} \right) = \dfrac{\pi }{2} + 2{\sin ^{ - 1}}x\] (ii)
We know, \[{\sin ^{ - 1}}z = \theta \] can be written as, \[z = \sin \theta \]. Using this concept in equation
(i), we get
\[\left( {1 - x} \right) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)\] (iii)
When sine function is of the form, \[\sin \left( {{{90}^ \circ } + \theta } \right)\] then we can write it as,
\[\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta \]
Using this for the term \[\sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)\] in equation (iii), we get
\[\left( {1 - x} \right) = \cos \left( {2{{\sin }^{ - 1}}x} \right)\] (iv)
Now, \[2{\sin ^{ - 1}}x\] can be written as, \[2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\]
Using this in equation (iii), we get
\[\left( {1 - x} \right) = \cos \left( {{{\cos }^{ - 1}}\left( {1 - 2{x^2}} \right)} \right)\]
\[ \Rightarrow \left( {1 - x} \right) = \left( {1 - 2{x^2}} \right)\]
\[ \Rightarrow - x = - 2{x^2}\]
\[ \Rightarrow 2{x^2} - x = 0\]
\[ \Rightarrow x\left( {2x - 1} \right) = 0\]
\[ \Rightarrow x = 0\,{\text{or}}\,\left( {2x - 1} \right) = 0\]
\[ \Rightarrow x = 0\,{\text{or}}\,x = \dfrac{1}{2}\]
We got two values for \[x\]. Now, we will find out whether both the values satisfy the given expression.
Putting \[x = 0\,\] in L.H.S of equation (i), we get,
\[L.H.S = {\sin ^{ - 1}}\left( {1 - 0} \right) - 2{\sin ^{ - 1}}0\]
\[L.H.S = {\sin ^{ - 1}}1\]
\[L.H.S = \dfrac{\pi }{2} = R.H.S\]
Therefore, \[x = 0\,\] satisfies the given expression.
Now, we check whether \[x = \dfrac{1}{2}\] satisfies the given expression. Putting \[x = \dfrac{1}{2}\] in the L.H.S of equation (i), we get
\[L.H.S = {\sin ^{ - 1}}\left( {1 - \dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\dfrac{1}{2}\]
\[ \Rightarrow L.H.S = {\sin ^{ - 1}}\dfrac{1}{2} - 2{\sin ^{ - 1}}\dfrac{1}{2}\]
\[ \Rightarrow L.H.S = - {\sin ^{ - 1}}\dfrac{1}{2}\]
\[ \Rightarrow L.H.S = \dfrac{\pi }{6}\]
Therefore here L.H.S is not equal to R.H.S, so \[x = \dfrac{1}{2}\] does not satisfy the given expression.
So, the only possible value for \[x\] is \[x = 0\].
Hence, the answer is \[x = 0\].

Note: There are three main functions in trigonometry, these are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the main functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.