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Question

Solve the following equation for x:

sin1(1x)2sin1x=π2, then x is equal to
a) 0,12
b) 1,12
c) zero
d) 12


Solution

Given

sin1(1x)2sin1x=π2  2sin1x=π2sin1(1x)2sin1x=cos1(1x)       [sin1(1x)+cos1(1x)=π2]
  cos(2sin1x)=1x    [multiplying both sides by cosx]
  cos(2sin1x)=1x      [cos(x)=+cos x]  [12sin2(sin1x)]=1x      [ cos2x=12sin2x]  12[sin(sin1x)]2=1x      [ sin2x=(sin x)2]  12x2=1x2x2x=0
  x(2x1)=0  x=0 or 2x1=0   x=0 or 12
But x=12 does not satisfy the given equation, so x = 0
Hence,the correct option is (c).

Alternate method.

Given, sin1(1x)2sin1x=π2
Let x=sinθθ=sin1x, then
sin1(1sin θ)2θ=π2sin1(1sinθ)=π2+2θ 1sinθ=sin(π2+2θ)1sinθ=cos2θ[ sin(π2+θ)=cosθ] 1sinθ=12sin2θ      [ cos2θ=12sin2θ] 2sin2θsinθ=0sinθ(2sinθ1)=0
Either sinθ or 2sinθ1=0
 x=0 or2x1=0      [sinθ=x] x=0 or x=12
But x=12 does not satisfy the given equation.  So, x=0
To check your answer
Put x=0 in the given equation,
 sin1(10)2sin10=π2π22×0=π2π2=π2
Put x=12 in the given equation,
 sin1(112)2sin112=π2sin2122sin112=π2π62×π6=π2π2π6=π2π6π2,so x=12 is not possibile

Note:  While solving the trigonometric equation, sometimes it may have same extra roots, so please take careful about it.
 

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