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
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
(i) sin 2θ=√32We have,sin 2θ=√32⇒sin 2θ=sin(π3)⇒2θ=nπ+(−1)nπ3,n∈zθ=nπ2+(−1)nπ6,n∈z(ii) cos 3θ=12We have,cos 3θ=12⇒cos 3θ=cos(π3)⇒ 3θ=2nπ±,n∈z⇒θ=2nπ9,n∈z(iii) sin 9θ=sinθsin 9θ−sinθ=0Apply sin A−sin B formulasinA−SinB=2 cos(A+B2)sin(A−B2)sin 9θ−sin θ=2cos5θ sin 4θ=0cos 5θ sin 4θ=0⇒cos 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 3θ⇒cos 3θ=sin 2θ⇒cos 3θ=cos (π4−2θ) [∵cos(π2−θ)=sinθ]⇒3θ=2nπ±(π2−2θ),n∈Z⇒either5θ=2nπ+π2,n∈z or θ=2nπ−π2,n∈z⇒5θ=(4n+1)π2,nπz or θ=(4n−1)π2⇒θ=(4n+1)π10,n∈z or θ(4n−1)π2,n∈z(v) tan θ+cot 2θ=0We have,tan θ+cot2θ=0tan θ=−cot2θ⇒cot2θ=−tanθ⇒1tan2θ=−1cotθ⇒tan 2θ=−cotθ⇒tan 2θ=−tan(π2−θ)⇒2θ=tan(θ−π2)⇒tan 2θ=nπ+(θ−π2),n∈z⇒θ=nπ−π2,n∈z(vi) tan 3θ+cot θ=0We have,tan 3θ+cotθ=0⇒tan 3θ=tan(π2−θ) [∵tan(π2−θ=cotθ)]⇒3θ=nπ+π2−θ,n∈z⇒4θ=nπ+π2−n∈z⇒θ=nπ4+π8,n∈z(vii) tan 2θ+tan θ 0=1We have,tan 2θ+cotθ 0=1⇒tan 2θ=1tanθ⇒tan 2θ=tan(π2−θ)⇒2θ=nπ+π2−θ,n∈z⇒3θ=nπ+π2−n∈z⇒θ=nπ3+π6,n∈z(viii) tan mθ+cot nθ=0sin mθ sin nθ+cos mθ cos nθ=0cos(m−n) θ=(2k+12)πθ=(2k+12(m−n))π,k∈Z(ix) tan pθ+cot qθWe have,tan pθ=cot qθ⇒tan pθ=tan(π2−qθ)⇒pθ=nπ±(π2−qθ),n∈z⇒(p+q) θ=(2n+1)π2,n∈z⇒θ=(2n+1)(p+q)π2,n∈z(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=(4m−1)π2or sin x=−12x=(4m−1)π2or x=(4n−1)π6,m,n∈z(xi) sin θ=tan θWe have,⇒sin θ=sinθcosθ⇒sin θ=sinθcosθ=0⇒sin θ(cos θ−1)=0⇒either sin θ=0 or cosθ−1=0⇒θ=nπ,n∈z or cos θ=1 ⇒cos θ=cos0∘Thus,θ=nθn∈z or θ=2nπ,m∈z(xii) sin 3θ+cos 2θ=0cos (2θ)=−sin (3θ)=−cos(π2−3θ)=cos(π2+3θ)⇒2n π+2θ=π2+3θθ=(4m−1)π2,m∈ZOr⇒2n π−2θ=π2+3θθ=(4n−1)π10,n∈Z
(i) sin 2θ=√32We have,sin 2θ=√32⇒sin 2θ=sin(π3)⇒2θ=nπ+(−1)nπ3,n∈zθ=nπ2+(−1)nπ6,n∈z(ii) cos 3θ=12We have,cos 3θ=12⇒cos 3θ=cos(π3)⇒ 3θ=2nπ±,n∈z⇒θ=2nπ9,n∈z(iii) sin 9θ=sinθsin 9θ−sinθ=0Apply sin A−sin B formulasinA−SinB=2 cos(A+B2)sin(A−B2)sin 9θ−sin θ=2cos5θ sin 4θ=0cos 5θ sin 4θ=0⇒cos 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 3θ⇒cos 3θ=sin 2θ⇒cos 3θ=cos (π4−2θ) [∵cos(π2−θ)=sinθ]⇒3θ=2nπ±(π2−2θ),n∈Z⇒either5θ=2nπ+π2,n∈z or θ=2nπ−π2,n∈z⇒5θ=(4n+1)π2,nπz or θ=(4n−1)π2⇒θ=(4n+1)π10,n∈z or θ(4n−1)π2,n∈z(v) tan θ+cot 2θ=0We have,tan θ+cot2θ=0tan θ=−cot2θ⇒cot2θ=−tanθ⇒1tan2θ=−1cotθ⇒tan 2θ=−cotθ⇒tan 2θ=−tan(π2−θ)⇒2θ=tan(θ−π2)⇒tan 2θ=nπ+(θ−π2),n∈z⇒θ=nπ−π2,n∈z(vi) tan 3θ+cot θ=0We have,tan 3θ+cotθ=0⇒tan 3θ=tan(π2−θ) [∵tan(π2−θ=cotθ)]⇒3θ=nπ+π2−θ,n∈z⇒4θ=nπ+π2−n∈z⇒θ=nπ4+π8,n∈z(vii) tan 2θ+tan θ 0=1We have,tan 2θ+cotθ 0=1⇒tan 2θ=1tanθ⇒tan 2θ=tan(π2−θ)⇒2θ=nπ+π2−θ,n∈z⇒3θ=nπ+π2−n∈z⇒θ=nπ3+π6,n∈z(viii) tan mθ+cot nθ=0sin mθ sin nθ+cos mθ cos nθ=0cos(m−n) θ=(2k+12)πθ=(2k+12(m−n))π,k∈Z(ix) tan pθ+cot qθWe have,tan pθ=cot qθ⇒tan pθ=tan(π2−qθ)⇒pθ=nπ±(π2−qθ),n∈z⇒(p+q) θ=(2n+1)π2,n∈z⇒θ=(2n+1)(p+q)π2,n∈z(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=(4m−1)π2or sin x=−12x=(4m−1)π2or x=(4n−1)π6,m,n∈z(xi) sin θ=tan θWe have,⇒sin θ=sinθcosθ⇒sin θ=sinθcosθ=0⇒sin θ(cos θ−1)=0⇒either sin θ=0 or cosθ−1=0⇒θ=nπ,n∈z or cos θ=1 ⇒cos θ=cos0∘Thus,θ=nθn∈z or θ=2nπ,m∈z(xii) sin 3θ+cos 2θ=0cos (2θ)=−sin (3θ)=−cos(π2−3θ)=cos(π2+3θ)⇒2n π+2θ=π2+3θθ=(4m−1)π2,m∈ZOr⇒2n π−2θ=π2+3θθ=(4n−1)π10,n∈Z
