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Question

# Solve : sin−1(1−x)−2sin−1x=π2, then x is equal to

A
0,12
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B
1,12
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C
0
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D
12
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Solution

## The correct option is B 0sin−1(1−x)−2sin−1x=π2⇒−2sin−1x=π2−sin−1(1−x)⇒−2sin−1x=cos−1(1−x).....(1)Let sin−1x=θ⇒sinθ=x⇒cosθ=√1−x2∴θ=cos−1(√1−x2)∴sin−1x=cos−1(√1−x2)Therefore, from equation (1), we have−2cos−1(√1−x2)=cos−1(1−x)Put x=siny−2cos−1(√1−sin2y)=cos−1(1−siny)⇒−2cos−1(cosy)=cos−1(1−siny)⇒−2y=cos−1(1−siny)⇒1−siny=cos(−2y)=cos2y=1−2sin2y⇒2sin2y−siny=0⇒siny(2siny−1)=0⇒siny=0or12∴x=0orx=12But, when x=12, it can be observed that :L.H.S.=sin−1(1−12)−2sin−112=sin−1(12)−2sin−112=−sin−112=−π6≠π2≠R.H.S.∴x=12 is not the solution of the given equation.Thus , x=0.Hence, the correct answer is C.

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