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Question

# The solutions of the equation sinx+3sin2x+sin3x=cosx+3cos2x+cos3x in the interval 0≤x≤2π are

A
π8,5π8,2π3
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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 π8,5π8,9π8,13π8sinx+3sin2x+sin3x=cosx+3cos2x+cos3x Using sinC+sinD=2sin(C+D2)cos(C−D2)And cosC+cosD=2cos(C+D2)cos(C−D2) we get2sin(x+3x2)cos(x−3x2)+3sin2x=2cos(x+3x2)cos(x−3x2)+3cos2x⇒2sin2xcosx+3sin2x=2cos2xcosx−3cos2xsin2x(2cosx+3)=cos2x(2cosx+3)⇒(sin2x−cos2x)(2cosx+3)=0⇒sin2x−cos2x=0⇒cos(π4+2x)=0⇒π4+2x=2nπ±π2⇒x=nπ±π4−π8 ...(1)Or 2cosx+3=0⇒cosx=−32No solution as −1≤cosx≤1∴ For 0≤x≤2πx=π8,5π8,9π8,13π8

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