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Question

The number of solutions of sin3x2cos3x22+sinx=cosx3 in the interval [0,10π] is

A
2
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B
4
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C
6
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D
5
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Solution

The correct option is D 5
Given, sin3x2cos3x22+sinx=cosx3

(sinx2cosx2)(1+sinx2cosx2)2(1+sinx2cosx2)=cos2x2sin2x233(sinx2cosx2)+2(sin2x2cos2x2)=0(sinx2cosx2)(3+2sinx2+2cosx2)=0sinx2=cosx2(sinθ+cosθ[2,2])tanx2=1x2=nπ+π4, nZx=2nπ+π2

In [0,10π],
x=π2,5π2,9π2,13π2,17π2
Number of solutions is 5.

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