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Question

# The solutions of the equation sinx−3sin2x+sin3x=cosx−3cos2x+cos3x in interval 0≤x≤2π 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 π8,5π8,9π8,13π8sinx−3sin2x+sin3x=cosx−3cos2x+cos3x⇒(sinx+sin3x)−3sin2x−(cosx+cos3x)+3cos2x=0⇒2sin(x+3x2)cos(x−3x2)−3sin2x−2cos(x+3x2)cos(x−3x2)+3cos2x=0⇒2sin2xcosx−3sin2x−2cos2xcosx+3cos2x=0⇒sin2x(2cosx−3)−cos2x(2cosx−3)=0⇒(2cosx−3)(sin2x−cos2x)=0⇒cosx=32 or sin2x=cos2xAs cosx∈[−1,1]Thus,cosx≠32So, sin2x=cos2x⇒tan2x=1⇒tan2x=tanπ4⇒2x=nπ+π4⇒x=nπ2+π8For n=0,x=π8For n=1,x=π2+π8=4π+π8=5π8For n=2,x=π+π8=9π8For n=3,x=3π2+π8=12π+π8=13π8Thus, the solutions are {π8,5π8,9π8,13π8}

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