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Question

The number of real solutions of the equation 1cos2x=2sin1(sinx),π<xπ, is:

A
\N
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B
1
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C
2
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D
infinite
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Solution

The correct option is C 2
2(sin2x)=2sin1sinx,π<xπ
(i)0xπ2, then sin x=sin1sinx=x.
X = 0 is obviously a solution and in 0<xπ2,sinx<x.
(ii)π2<xπ, we get
sinx=sin1sin(πx)=πx
x=π is only solution in this case.
(iii)π2x<0,sinx=x which has no solution in this interval.
(iv)π<x<π2sinx=πxsinx=π+x
Which has no solution in the given interval.
Hence, there are only two solutions x=0,π

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