Solve the following equations :(i) cos θ+cps 2θ+cos 3θ=0(ii) cos θ+cos 3θ−cos 2θ=0(iii) sin θ+sin 5θ=sin 3θ(iv) cos θ cos 2θ cos 3θ=14(v) cos θ+sin θ=cos 2θ+sin2θ(vi) sin θ+sin 2θ+sin 3θ=0(vii) sin θ+sin 2θ+sin 3θ+sin 4θ=0(viii) sin 3θ−sin θ=4 cos2θ−2(ix) sin 2θ−sin 4θ+sin 6θ=0
Solve the following equations :(i) cos θ+cps 2θ+cos 3θ=0(ii) cos θ+cos 3θ−cos 2θ=0(iii) sin θ+sin 5θ=sin 3θ(iv) cos θ cos 2θ cos 3θ=14(v) cos θ+sin θ=cos 2θ+sin2θ(vi) sin θ+sin 2θ+sin 3θ=0(vii) sin θ+sin 2θ+sin 3θ+sin 4θ=0(viii) sin 3θ−sin θ=4 cos2θ−2(ix) sin 2θ−sin 4θ+sin 6θ=0
(i) cos θ+cps 2θ+cos 3θ=0⇒cos 2θ+2cos 2θ.cos θ=0[∵cos θ+cos 3θ=2 cos 2θ.cos θ]⇒cos 2θ(1+2cos θ)=0eithercos 2θ=0 Or 1+2 cos θ=0⇒2θ=(2n+1)π4,n∈z Or cos θ=−12⇒θ=(2n+1)π4,n∈z or cos θ=+cos(π−π3) Or cos θ=cos 2π3 Or θ=2nπ±2π3,n∈zThus,θ=(2n+1)π4, Or(2nπ±2π3),n∈z(ii) cos θ+cos 3θ−cos 2θ=0⇒2 cos 2θ.cos θ−cos 2θ=0⇒cos 2θ(2cos θ−1)=0eithercos 2θ=0 Or 2 cos θ=1⇒2θ=(2n+1)π2,n∈zOr cos θ=12=cos π3⇒θ=(2n+1)π4,n∈zOr θ=2mπ±π3,m∈z(iii) sin θ+sin 5θ=sin 3θ⇒2sin 3θ.cos 2θ−sin 3θ=0[∵sin C+sin D=2 sin C+D2,cosC−D2]⇒sin 3θ[2 cos 2θ−1]eithersin 3θ=0 Or 2 cos−1=0⇒3θ=nπ,n∈z Or cos 2θ=12=cosπ3⇒θ=nπ3,n∈z Or 2θ=2mπ±π3,m∈z Or θ=mπ±π6Thus,θ=nπ3 Or mπ±π6,n,m,∈z(iv) cos θ cos 2θ cos 3θ=14We have,cos θ cos 2θ cos 3θ=14⇒2 cos θ.cos 3θ.cos 2θ=12⇒(cos 4θ+cos 2θ)cos 2θ=12⇒(2cos2 2θ−1+cos 2θ)cos 2θ=12⇒2cos3 2θ+cos2 2θ−cos 2θ=12⇒4cos2 2θ+2 cos2 2θ−2 cos 2θ−1=0⇒(2cos2 θ−1)(2cos 2θ+1)=0eitherOr ⇒2cos 3θ+1=0⇒cos 4θ=0 Or ⇒cos 2θ=−12⇒4θ=(2n+1)π2 Or⇒cos 2θ=cos 2π3⇒θ=(2n+1)π8 Or⇒2θ=2mπ±2π3⇒θ=mπ±π3Thus,θ=(2n+1)π8Or θ=mπ±π3,m,n∈z(v) cos θ+sin θ=cos 2θ+sin2θWe have,cos θ+sin θ=cos 2θ+sin2θ⇒cos θ−cos 2θ=sin 2θ−sin2θ⇒2 sin302,sinθ2=2cos3θ2,sinθ2⇒2 sin θ2 (sin3θ2−cos3θ2)=0eithersin θ2=0 Or sin 3θ2− cos 3θ2=0⇒θ2=nπ,n∈zOr tan 3θ2=1 tan π4⇒θ=2nπ,n∈z Or3θ2=nπ+π4Or θ =2n π3+π32,n∈zThus,⇒θ=2mπ Or 2nπ3+π6,n∈z(vi) sin θ+sin 2θ+sin 3θ=0We have,sin θ+sin 2θ+sin 3θ=0⇒sin 2θ+2 sin 2θ.cos θ=0⇒sin 2θ+(1+2 cos θ)=0⇒eithersin 2θ=0 Or 1+2 cos θ=0⇒2θ=nπ,n∈zOr cosθ=−12=cos(π−π3)⇒θ=nπ2,n∈zOr θ=2mπ± 2π3,m∈zThus,θ=nπ2,n∈z Or θ=2mπ±2π3,m∈z(vii) sin θ+sin 2θ+sin 3θ+sin 4θ=0Given:sinx+sin x+sin 3x+sin 4x=0(sin 4θ+sin 2θ)+(sin 3θ+sin θ)=0Using,(sin A+sin B)formula=>2 sin [(4θ+2θ)2] cos [(4θ−2θ)2]+2 sin [(3θ+θ)2] cos [(3θ−θ)2]=02 sin3θ cosθ+2 sin 2θ cos θ=02 cos θ(sin 3θ+sin 2θ)=02 cos θ(2 sin [(3θ+2θ)2] cos [(3θ−2θ)2])=04 cos θ sin 5θ2 cos π2=0cos θ=0;sin5θ2=0;cosπ2=0θ=(2n+1)π2;5θ2=mπ;θ2(2r+1)π2θ=(2n+1)π2;θ=2mπ5;θ=(2r+1)π,m,r,n∈z(viii) sin 3θ−sin θ=4 cos2θ−2We have,sin 3θ−sin θ=4 cos2θ−2⇒2 cos 2θ.sin θ=2 (2 cos2θ−1)⇒2 cos 2θ.sin θ=2 cos 2θ [∵ cos 2θ=2 cos2θ−1]⇒2 cos 2θ(sin θ−1)=0eithercos 2θ=0 Or sin θ−1=0⇒2θ=(2n+1)π2,n∈zOr sin θ=1=sin π2⇒θ=(2n+1)π4,n∈zOr θ=mπ+(−1)mπ2,m∈zThus,θ=(2n+1)π4,n∈zor mπ+(−1)Mπ2,m∈z(ix) sin 2θ−sin 4θ+sin 6θ=0sin 2θ−sin 4θ+sin 6θ=0(sin 2θ+sin 6θ)−sin 4θ=02 sin(802),cos (402)−sin 4θ=02 sin 4θ.cos 2θ−sin 4θ=0sin 4θ(2 cos 2θ−1)=0sin 4θ=0 or 2 cos 2θ−1=04θ=n(π) or cos 2θ=1/2θ=(nπ4)or cos 2θ=cos(π3)θ=(nπ4)or θ=n(π)(π6)
(i) cos θ+cps 2θ+cos 3θ=0⇒cos 2θ+2cos 2θ.cos θ=0[∵cos θ+cos 3θ=2 cos 2θ.cos θ]⇒cos 2θ(1+2cos θ)=0eithercos 2θ=0 Or 1+2 cos θ=0⇒2θ=(2n+1)π4,n∈z Or cos θ=−12⇒θ=(2n+1)π4,n∈z or cos θ=+cos(π−π3) Or cos θ=cos 2π3 Or θ=2nπ±2π3,n∈zThus,θ=(2n+1)π4, Or(2nπ±2π3),n∈z(ii) cos θ+cos 3θ−cos 2θ=0⇒2 cos 2θ.cos θ−cos 2θ=0⇒cos 2θ(2cos θ−1)=0eithercos 2θ=0 Or 2 cos θ=1⇒2θ=(2n+1)π2,n∈zOr cos θ=12=cos π3⇒θ=(2n+1)π4,n∈zOr θ=2mπ±π3,m∈z(iii) sin θ+sin 5θ=sin 3θ⇒2sin 3θ.cos 2θ−sin 3θ=0[∵sin C+sin D=2 sin C+D2,cosC−D2]⇒sin 3θ[2 cos 2θ−1]eithersin 3θ=0 Or 2 cos−1=0⇒3θ=nπ,n∈z Or cos 2θ=12=cosπ3⇒θ=nπ3,n∈z Or 2θ=2mπ±π3,m∈z Or θ=mπ±π6Thus,θ=nπ3 Or mπ±π6,n,m,∈z(iv) cos θ cos 2θ cos 3θ=14We have,cos θ cos 2θ cos 3θ=14⇒2 cos θ.cos 3θ.cos 2θ=12⇒(cos 4θ+cos 2θ)cos 2θ=12⇒(2cos2 2θ−1+cos 2θ)cos 2θ=12⇒2cos3 2θ+cos2 2θ−cos 2θ=12⇒4cos2 2θ+2 cos2 2θ−2 cos 2θ−1=0⇒(2cos2 θ−1)(2cos 2θ+1)=0eitherOr ⇒2cos 3θ+1=0⇒cos 4θ=0 Or ⇒cos 2θ=−12⇒4θ=(2n+1)π2 Or⇒cos 2θ=cos 2π3⇒θ=(2n+1)π8 Or⇒2θ=2mπ±2π3⇒θ=mπ±π3Thus,θ=(2n+1)π8Or θ=mπ±π3,m,n∈z(v) cos θ+sin θ=cos 2θ+sin2θWe have,cos θ+sin θ=cos 2θ+sin2θ⇒cos θ−cos 2θ=sin 2θ−sin2θ⇒2 sin302,sinθ2=2cos3θ2,sinθ2⇒2 sin θ2 (sin3θ2−cos3θ2)=0eithersin θ2=0 Or sin 3θ2− cos 3θ2=0⇒θ2=nπ,n∈zOr tan 3θ2=1 tan π4⇒θ=2nπ,n∈z Or3θ2=nπ+π4Or θ =2n π3+π32,n∈zThus,⇒θ=2mπ Or 2nπ3+π6,n∈z(vi) sin θ+sin 2θ+sin 3θ=0We have,sin θ+sin 2θ+sin 3θ=0⇒sin 2θ+2 sin 2θ.cos θ=0⇒sin 2θ+(1+2 cos θ)=0⇒eithersin 2θ=0 Or 1+2 cos θ=0⇒2θ=nπ,n∈zOr cosθ=−12=cos(π−π3)⇒θ=nπ2,n∈zOr θ=2mπ± 2π3,m∈zThus,θ=nπ2,n∈z Or θ=2mπ±2π3,m∈z(vii) sin θ+sin 2θ+sin 3θ+sin 4θ=0Given:sinx+sin x+sin 3x+sin 4x=0(sin 4θ+sin 2θ)+(sin 3θ+sin θ)=0Using,(sin A+sin B)formula=>2 sin [(4θ+2θ)2] cos [(4θ−2θ)2]+2 sin [(3θ+θ)2] cos [(3θ−θ)2]=02 sin3θ cosθ+2 sin 2θ cos θ=02 cos θ(sin 3θ+sin 2θ)=02 cos θ(2 sin [(3θ+2θ)2] cos [(3θ−2θ)2])=04 cos θ sin 5θ2 cos π2=0cos θ=0;sin5θ2=0;cosπ2=0θ=(2n+1)π2;5θ2=mπ;θ2(2r+1)π2θ=(2n+1)π2;θ=2mπ5;θ=(2r+1)π,m,r,n∈z(viii) sin 3θ−sin θ=4 cos2θ−2We have,sin 3θ−sin θ=4 cos2θ−2⇒2 cos 2θ.sin θ=2 (2 cos2θ−1)⇒2 cos 2θ.sin θ=2 cos 2θ [∵ cos 2θ=2 cos2θ−1]⇒2 cos 2θ(sin θ−1)=0eithercos 2θ=0 Or sin θ−1=0⇒2θ=(2n+1)π2,n∈zOr sin θ=1=sin π2⇒θ=(2n+1)π4,n∈zOr θ=mπ+(−1)mπ2,m∈zThus,θ=(2n+1)π4,n∈zor mπ+(−1)Mπ2,m∈z(ix) sin 2θ−sin 4θ+sin 6θ=0sin 2θ−sin 4θ+sin 6θ=0(sin 2θ+sin 6θ)−sin 4θ=02 sin(802),cos (402)−sin 4θ=02 sin 4θ.cos 2θ−sin 4θ=0sin 4θ(2 cos 2θ−1)=0sin 4θ=0 or 2 cos 2θ−1=04θ=n(π) or cos 2θ=1/2θ=(nπ4)or cos 2θ=cos(π3)θ=(nπ4)or θ=n(π)(π6)
