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Question Number 44920 by peter frank last updated on 06/Oct/18
solve.3sin2x+2cos2x+31−sin2x+2sin2x=28
Answered by ajfour last updated on 06/Oct/18
3sin2x+2cos2x+31−sin2x+2sin2x=28max(sin2x+2cos2x)=max(sin2x+cos2x+1)=2+1>2max(1−sin2x+2sin2x)=max(1−sin2x+1−cos2x)=2+2>328=30+33thisiscertainlypossibleforcos2x=−1⇒2x=2nπ+π⇒x=nπ+π2∀x∈Z.
Answered by hknkrc46 last updated on 24/Dec/18
3sin2x+2cos2x+31−sin2x+2sin2x=283sin2x+(cos2x+1)+31−sin2x+(1−cos2x)3(sin2x+cos2x)+1+32−(sin2x+cos2x)⇒sin2x+cos2x=−1⇒2sinxcosx+cos2x−sin2x=−cos2x−sin2x⇒2cos2x+2sinxcosx=0⇒cosx(cosx+sinx)=0⇒cosx=0∨cosx+sinx=0⇒cosx=0∨(cosx+sinx)2=0⇒cosx=0∨1+sin2x=0⇒cosx=0∨sin2x=−1⇒x=(4k+34)π,k∈Z
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