All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 4429 by alib last updated on 24/Jan/16
(sinx+3cosx)sin3x=2
Answered by Yozzii last updated on 24/Jan/16
sinx+3cosx=1+3sin(x+π/3)=2sin(x+π3)(sinx+3cosx)sin3x=2.........(Υ)∴2sin(x+π/3)sin3x=2sin(x+π/3)sin3x=1−0.5(cos(4x+π/3)−cos(2x−π/3))=1cos(4x+π/3)−cos(2x−π/3)=−2(∗)(∗)istrueiffcos(4x+π/3)=−1andcos(2x−π/3)=1.Fromcos(4x+π/3)=−1⇒4x+π/3=2nπ±π(n∈Z)4x=(2n−13±1)πx=6n−1±312π⇒x=(3n−2)π6orx=(3n+1)π6.Fromcos(2x−π/3)=1⇒2x−π/3=2rπ⇒x=12(2rπ+13)=(6r+1)π6=(r+16)π(r∈Z)LetX1={x∈R,n∈Z∣x=(3n−2)π6},X2={x∈R,n∈Z∣x=(3n+1)π6},X3={x∈R,r∈Z∣x=(6r+1)π6}.Thesearethethreesolutionsetsobtainedabove.qmayormaynotequaln.Wemustsearchforasetofvaluesofxthatsatisfiesbothequations.So,weinvestigatewhenX1=X3andwhenX2=X3.IfX1=X3⇒3n−2=6r+1⇒3(n−2r)=3⇒n=2r+1∴n=2r+1(oddn).Sofor(Υ),sin3x=sin(3(π(6r+1)6))=sin(π(6r+12))=sin(3πr+π2)=sin3πrcos(0.5π)+cos3πrsin0.5πsin3x=(−1)r2sin(x+π3)=2sin(πr+π6+π3)=2sin(πr+3+618π)=2sin(πr+0.5π)=2(sinπrcos0.5π+sin0.5πcosπr)=2(−1)r∴2(−1)r(−1)r=2(−1)2r=2X3canbeusedonitsownasonegeneralsolutionsetcoveringX1.IfX2=X3⇒3n+1=6r+1⇒n=2r(evenn).Now,Z=O∪E.Sincen∈Z⇒n∈Oorn∈E,OandEbeingdisjoint.SinceX1=X3foroddnandX2=X3forevenn,forallr∈Zineachcase,thenX3=X1∪X2;i.eX3constitutesallsolutionsof(Υ).
Terms of Service
Privacy Policy
Contact: info@tinkutara.com