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Question

If x1 and x2 are the two solutions of the equation 2cosxsin3x=sin4x+1 lying in the interval [0,2π] then the value of |x1x2| is

A
π4
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B
π2
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C
π
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D
2π
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Solution

The correct option is C π
If x1&x2ax()of(1)in[0,2π]
2cosxsin3x=sin4x+1
sin4x+sin2x=sin4x+1
2cosxsiny=sin(x+y)sin(xy)
sin2x=1
sin2x=sinπ2
2x=π2+2nπ
x=π4+nπ
x=π4+nπ
x1=π4+π
x2=π4
|x2x1|=π4+ππ4=π

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