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

If sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}, then xx equals.

Explanation

Solution

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

Complete step by step solution:
Given, the equation sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2} (i)
To find the value of xx let us simplify the equation further,
sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}
⇒sin⁡−1(1−x)=π2+2sin⁡−1x\Rightarrow {\sin ^{ - 1}}\left( {1 - x} \right) = \dfrac{\pi }{2} + 2{\sin ^{ - 1}}x (ii)
We know, sin⁡−1z=θ{\sin ^{ - 1}}z = \theta can be written as, z=sin⁡θz = \sin \theta . Using this concept in equation
(i), we get
(1−x)=sin⁡(π2+2sin⁡−1x)\left( {1 - x} \right) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right) (iii)
When sine function is of the form, sin⁡(90∘+θ)\sin \left( {{{90}^ \circ } + \theta } \right) then we can write it as,
sin⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta
Using this for the term sin⁡(π2+2sin⁡−1x)\sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right) in equation (iii), we get
(1−x)=cos⁡(2sin⁡−1x)\left( {1 - x} \right) = \cos \left( {2{{\sin }^{ - 1}}x} \right) (iv)
Now, 2sin⁡−1x2{\sin ^{ - 1}}x can be written as, 2sin⁡−1x=cos⁡−1(1−2x2)2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)
Using this in equation (iii), we get
(1−x)=cos⁡(cos⁡−1(1−2x2))\left( {1 - x} \right) = \cos \left( {{{\cos }^{ - 1}}\left( {1 - 2{x^2}} \right)} \right)
⇒(1−x)=(1−2x2)\Rightarrow \left( {1 - x} \right) = \left( {1 - 2{x^2}} \right)
⇒−x=−2x2\Rightarrow - x = - 2{x^2}
⇒2x2−x=0\Rightarrow 2{x^2} - x = 0
⇒x(2x−1)=0\Rightarrow x\left( {2x - 1} \right) = 0
⇒x=0 or (2x−1)=0\Rightarrow x = 0\,{\text{or}}\,\left( {2x - 1} \right) = 0
⇒x=0 or x=12\Rightarrow x = 0\,{\text{or}}\,x = \dfrac{1}{2}
We got two values for xx. Now, we will find out whether both the values satisfy the given expression.
Putting x=0 x = 0\, in L.H.S of equation (i), we get,
L.H.S=sin⁡−1(1−0)−2sin⁡−10L.H.S = {\sin ^{ - 1}}\left( {1 - 0} \right) - 2{\sin ^{ - 1}}0
L.H.S=sin⁡−11L.H.S = {\sin ^{ - 1}}1
L.H.S=π2=R.H.SL.H.S = \dfrac{\pi }{2} = R.H.S
Therefore, x=0 x = 0\, satisfies the given expression.
Now, we check whether x=12x = \dfrac{1}{2} satisfies the given expression. Putting x=12x = \dfrac{1}{2} in the L.H.S of equation (i), we get
L.H.S=sin⁡−1(1−12)−2sin⁡−112L.H.S = {\sin ^{ - 1}}\left( {1 - \dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\dfrac{1}{2}
⇒L.H.S=sin⁡−112−2sin⁡−112\Rightarrow L.H.S = {\sin ^{ - 1}}\dfrac{1}{2} - 2{\sin ^{ - 1}}\dfrac{1}{2}
⇒L.H.S=−sin⁡−112\Rightarrow L.H.S = - {\sin ^{ - 1}}\dfrac{1}{2}
⇒L.H.S=π6\Rightarrow L.H.S = \dfrac{\pi }{6}
Therefore here L.H.S is not equal to R.H.S, so x=12x = \dfrac{1}{2} does not satisfy the given expression.
So, the only possible value for xx is x=0x = 0.
Hence, the answer is x=0x = 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.