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Question Number 165070 by mathlove last updated on 25/Jan/22
Answered by bobhans last updated on 25/Jan/22
cosx−sinx=1−2sinxcosx⇒1−2sinxcosxsinxcosx=221−2t=22t,[t=sinxcosx]⇒1−2t=8t2⇒8t2+2t−1=0⇒(4t−1)(2t+1)=0⇒{t=sinxcosx=14;sin2x=12t=sinxcosx=−12;sin2x=−1
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