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Question Number 144333 by gsk2684 last updated on 24/Jun/21
findthenumberofsolutionsof1+sinx.sin2x2=0in[−ΠΠ]
Commented by MJS_new last updated on 25/Jun/21
nosolution:1−338⩽1+sinxsin2x2⩽1+338
lett=tanx2⇒sinx=2tt2+1∧sinx2=tt2+11+sinxsin2x2=t4+2t3+2t2+1(t2+1)2ddt[t4+2t3+2t2+1(t2+1)2]=0−2t2(t2−3)(t2+1)3=0⇒t=0∨t=±3⇒1−338⩽t4+2t3+2t2+1(t2+1)2⩽1+338
Commented by gsk2684 last updated on 26/Jun/21
thankyouMJS.
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