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Question Number 136121 by bramlexs22 last updated on 18/Mar/21
3cosx=13sin(2x3)+17cos(x3)
Answered by EDWIN88 last updated on 19/Mar/21
letx3=t⇒3cos3t=13sin2t+17cost⇒3(4cos3t−3cost)=26sintcost+17cost12cos3t−26cost−26sintcost=02cost(6cos2t−13sint−13)=0(∗)2cost=0⇒t=±π2+2nπx3=±π2+2nπ⇒x=±3π2+6nπ(∗)⇒6−6sin2t−13sint−13=0⇒6sin2t+13sint+7=0⇒(6sint+7)(sint+1)=0{sint=−76⇒t∉Rsint=−1⇒t=−π2+2nπ,x3=−π2+2nπ
Commented by EDWIN88 last updated on 19/Mar/21
yes.thanksyou
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