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Question

# The general solution of equationsinx−3sin2x+sin3x=cosx−3cos2x+cos3x is-

A
nπ+π8
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B
nπ2+π8
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C
2nπ+cos1(13)
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D
2nπ+cos1(32)
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Solution

## The correct option is B nπ2+π8sinx−3sin2x+sin3x=cosx−3cos2x+cos3x⇒sinx−6sinxcosx+3sinx−4sin2x=cosxGivensinx−3sin2x+sin3x=cosx−3cos2x+cos3x⇒sin3x+sinx−3sin2x=cos3x+cosx−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒2sin2xcosx−2cos2xcosx=3sin2x−3cos2x⇒2cosx(sin2x−cos2x)=3(sin2x−cos2x)⇒(2cosx−3)(sin2x−cos2x)=0∴cosx=32, or sinx=cos2x↓Impossible ⇒tan2x=1⇒tan2x=tanπ4⇒2x=xπ+π4⇒x=xπ2+π8Hence, Genral solution is x=xπ2+π8

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