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Question

The solutions of the equation sinx3sin2x+sin3x=cosx3cos2x+cos3x in interval 0x2π are :

A
π8,5π8,9π8
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B
π8,5π8,9π8,13π8
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C
4π3,9π3,2π3,13π8
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D
π8,5π8,9π3,4π3
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Solution

The correct option is B π8,5π8,9π8,13π8
sinx3sin2x+sin3x=cosx3cos2x+cos3x

(sinx+sin3x)3sin2x(cosx+cos3x)+3cos2x=0

2sin(x+3x2)cos(x3x2)3sin2x2cos(x+3x2)cos(x3x2)+3cos2x=0

2sin2xcosx3sin2x2cos2xcosx+3cos2x=0

sin2x(2cosx3)cos2x(2cosx3)=0

(2cosx3)(sin2xcos2x)=0

cosx=32 or sin2x=cos2x

As cosx[1,1]

Thus,cosx32

So, sin2x=cos2x

tan2x=1

tan2x=tanπ4

2x=nπ+π4

x=nπ2+π8

For n=0,x=π8

For n=1,x=π2+π8=4π+π8=5π8

For n=2,x=π+π8=9π8

For n=3,x=3π2+π8=12π+π8=13π8

Thus, the solutions are {π8,5π8,9π8,13π8}

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