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Question

Find the general solutions of the following equations:(i) sin 2θ=32(ii) cos 3θ=12(iii) sin 9θ=sinθ(iv) sin θ=cos 3θ(v) tan θ+cot 2θ=0(vi) tan 3θ=cot θ(vii) tan 2θ tan θ=1(viii) tan mθ+cot nθ=0(ix) tan pθ=cot qθ(x) sin 2θ+cos θ=0(xi)sin θ=tan θ(xii)sin 3θ+cos 2θ=0

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Solution

(i) sin 2θ=32We have,sin 2θ=32sin 2θ=sin(π3)2θ=nπ+(1)nπ3,nzθ=nπ2+(1)nπ6,nz(ii) cos 3θ=12We have,cos 3θ=12cos 3θ=cos(π3) 3θ=2nπ±,nzθ=2nπ9,nz(iii) sin 9θ=sinθsin 9θsinθ=0Apply sin Asin B formulasinASinB=2 cos(A+B2)sin(AB2)sin 9θsin θ=2cos5θ sin 4θ=0cos 5θ sin 4θ=0cos 5θ=0(or)4θ=05θ={(2n+1)π2}(or)4θ=nπθ={(2n+1)π2}(or)θ={nπ4}Where nπZ(iv) sin 2θ=cos 3θWe have,sin 2θ==cos 3θcos 3θ=sin 2θcos 3θ=cos (π42θ) [cos(π2θ)=sinθ]3θ=2nπ±(π22θ),nZeither5θ=2nπ+π2,nz or θ=2nππ2,nz5θ=(4n+1)π2,nπz or θ=(4n1)π2θ=(4n+1)π10,nz or θ(4n1)π2,nz(v) tan θ+cot 2θ=0We have,tan θ+cot2θ=0tan θ=cot2θcot2θ=tanθ1tan2θ=1cotθtan 2θ=cotθtan 2θ=tan(π2θ)2θ=tan(θπ2)tan 2θ=nπ+(θπ2),nzθ=nππ2,nz(vi) tan 3θ+cot θ=0We have,tan 3θ+cotθ=0tan 3θ=tan(π2θ) [tan(π2θ=cotθ)]3θ=nπ+π2θ,nz4θ=nπ+π2nzθ=nπ4+π8,nz(vii) tan 2θ+tan θ 0=1We have,tan 2θ+cotθ 0=1tan 2θ=1tanθtan 2θ=tan(π2θ)2θ=nπ+π2θ,nz3θ=nπ+π2nzθ=nπ3+π6,nz(viii) tan mθ+cot nθ=0sin mθ sin nθ+cos mθ cos nθ=0cos(mn) θ=(2k+12)πθ=(2k+12(mn))π,kZ(ix) tan pθ+cot qθWe have,tan pθ=cot qθtan pθ=tan(π2qθ)pθ=nπ±(π2qθ),nz(p+q) θ=(2n+1)π2,nzθ=(2n+1)(p+q)π2,nz(x) sin 2θ+cos θ=02sin x cos x +cos=0cos x(2 sin x+1)=0cos x=0 or 2 sin x+1=0x=(4m1)π2or sin x=12x=(4m1)π2or x=(4n1)π6,m,nz(xi) sin θ=tan θWe have,sin θ=sinθcosθsin θ=sinθcosθ=0sin θ(cos θ1)=0either sin θ=0 or cosθ1=0θ=nπ,nz or cos θ=1 cos θ=cos0Thus,θ=nθnz or θ=2nπ,mz(xii) sin 3θ+cos 2θ=0cos (2θ)=sin (3θ)=cos(π23θ)=cos(π2+3θ)2n π+2θ=π2+3θθ=(4m1)π2,mZOr2n π2θ=π2+3θθ=(4n1)π10,nZ


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