Solve the following equations:(i) sin2 θ−cos θ=14(ii) 2 cos2θ−5 cos θ+2=0(iii) 2 sin2x+√3cos x+1=0(iv) 4 sin2θ−8 cosθ+1=0(v) tan2 x+(1−√3)tan x−√3=0(vi) 3 cos2θ−2√3 sinθ cos θ−3 sin2θ=0(vii) cos 4θ=cos 2θ
Solve the following equations:(i) sin2 θ−cos θ=14(ii) 2 cos2θ−5 cos θ+2=0(iii) 2 sin2x+√3cos x+1=0(iv) 4 sin2θ−8 cosθ+1=0(v) tan2 x+(1−√3)tan x−√3=0(vi) 3 cos2θ−2√3 sinθ cos θ−3 sin2θ=0(vii) cos 4θ=cos 2θ
(i) sin2 θ−cos θ=14We have,sin2θ−cos θ=14⇒1−cos2 θ−cos θ=14 [∵sin2 θ=1−cos2 θ]⇒cos2 θ+cos θ−34⇒4cos2 θ+4cos θ−3=0⇒4cos2 θ+6cos θ−2cos θ−3=0 [factorize it]⇒2cosθ(2cos θ+3)−1(cos θ+3)=0⇒(2cos θ−1)(2cos θ+3)=0⇒either2cos θ−1=0 Or 2cos θ+3=0⇒cos θ=12 Or cos θ=−32[This is not possible as−1]⇒cos θ=cosπ3⇒θ=2nπ±π3,n∈Z(ii) 2 cos2θ−5 cos θ+2=0We have,2 cos2 θ−5 cos θ+2=0⇒2 cos2θ−4 cos θ−cos θ+2=0 [use factorization]⇒2 cos θ(cos θ−2)−1(cos θ−2)=0⇒(2cos θ−1)(cos θ−2)=0⇒either2 cos θ−1=0 Or cos θ=2⇒cos θ=cosπ3 [This is not possible as−1<cos θ<1]⇒θ=2nπ±π3,n∈zThus,θ=2nπ±π3,n∈z(iii) 2 sin2x+√3cos x+1=0We have,2 sn2 x+√3 cos x+1=0⇒2(1−cos2 x)+√3 cos x+1=0⇒2 cos2 x−√3 cos x−3=0factorise it we get,⇒2 cos2 x−2√3 cos x+√3 cos x−3=0⇒2 cos x(cos x−√3)+√3(cos x−√3)=0⇒(2cos x+√3)+(cos x−√3)=0⇒eithercos x=−√32=Or cos x=√3[This in not possible as−1<cos x<1]⇒cos x=cos(π−π6)⇒cos x=cos5π6⇒x=2nπ±5π6,n∈z(iv) 4 sin2θ−8 cosθ+1=0We have,4 sin2θ−8 cosθ+1=0⇒4(1−cos2 θ)−8cos θ+1=0⇒4 cos2 θ+8 cos θ−5=0factorise it,we get,⇒4 cos2 θ+10 cos θ−2 cos θ−5=0⇒2 cos θ(2 cos θ+5)−1(2 cos θ+5)=0⇒(2 cos θ−1)(2 cos θ+5)=0either 2 cos θ−1=0 Or 2 cos θ+5=0⇒cos θ=12 Or cos θ=−52[This is not possible as−1<cos θ<1]⇒cos θ=cosπ3⇒ θ=2nπ±π3,n∈z(v) tan2 x+(1−√3)tan x−√3=0We have,tan2 x+(1−√3)tan x−√3=0⇒tan2 x+tan x−√3 tan x−√3=0⇒tan x(tan x+1)−√3(tan x+1)=0⇒(tan x−√3)(tan x+1)=0⇒eithertan x=√3 Or tan x=−1⇒tan x=tan π3Or tan x=−tan π4⇒x=nπ+ π3,n∈z Or x=mπ−π4,m∈z∴x=nπ+ π3Or mπ−π4,n,m∈z(vi) 3 cos2θ−2√3 sinθ cos θ−3 sin2θ=0√3 cos2 θ−sin θ cos θ−√3 sin2 θ=0 (Divideni by√3)√3 cos2 θ+2 sin θ cos θ−3 sin θ cos θ−√3 sin2 θ=0cos θ(√3 cos θ+sin θ)−√3 sin θ(√3 cos θ+sin θ)=0(√3 cos θ+sin θ)(cos θ−√3 sin θ)=0√3 cos θ+sin θ=0Or cos θ−√3 sin θ=0tan θ=−√3=−tanπ3Or tan θ= 1√3=tanπ6 θ=nπ−π3 Or θ=mπ+π6n,m∈z(vii) cos 4θ=cos 2θWe have,cos 4θ=cos 2θ⇒cos 4θ−cos 2θ=0⇒2 sin θ.sin 3θ=0⇒eithersin θ=0 Or sin 3θ=0⇒θ=nπ,n∈z Or 3θ=mπ, m∈zThus,θ=nπ, Or m π3,n ,m ∈z
(i) sin2 θ−cos θ=14We have,sin2θ−cos θ=14⇒1−cos2 θ−cos θ=14 [∵sin2 θ=1−cos2 θ]⇒cos2 θ+cos θ−34⇒4cos2 θ+4cos θ−3=0⇒4cos2 θ+6cos θ−2cos θ−3=0 [factorize it]⇒2cosθ(2cos θ+3)−1(cos θ+3)=0⇒(2cos θ−1)(2cos θ+3)=0⇒either2cos θ−1=0 Or 2cos θ+3=0⇒cos θ=12 Or cos θ=−32[This is not possible as−1]⇒cos θ=cosπ3⇒θ=2nπ±π3,n∈Z(ii) 2 cos2θ−5 cos θ+2=0We have,2 cos2 θ−5 cos θ+2=0⇒2 cos2θ−4 cos θ−cos θ+2=0 [use factorization]⇒2 cos θ(cos θ−2)−1(cos θ−2)=0⇒(2cos θ−1)(cos θ−2)=0⇒either2 cos θ−1=0 Or cos θ=2⇒cos θ=cosπ3 [This is not possible as−1<cos θ<1]⇒θ=2nπ±π3,n∈zThus,θ=2nπ±π3,n∈z(iii) 2 sin2x+√3cos x+1=0We have,2 sn2 x+√3 cos x+1=0⇒2(1−cos2 x)+√3 cos x+1=0⇒2 cos2 x−√3 cos x−3=0factorise it we get,⇒2 cos2 x−2√3 cos x+√3 cos x−3=0⇒2 cos x(cos x−√3)+√3(cos x−√3)=0⇒(2cos x+√3)+(cos x−√3)=0⇒eithercos x=−√32=Or cos x=√3[This in not possible as−1<cos x<1]⇒cos x=cos(π−π6)⇒cos x=cos5π6⇒x=2nπ±5π6,n∈z(iv) 4 sin2θ−8 cosθ+1=0We have,4 sin2θ−8 cosθ+1=0⇒4(1−cos2 θ)−8cos θ+1=0⇒4 cos2 θ+8 cos θ−5=0factorise it,we get,⇒4 cos2 θ+10 cos θ−2 cos θ−5=0⇒2 cos θ(2 cos θ+5)−1(2 cos θ+5)=0⇒(2 cos θ−1)(2 cos θ+5)=0either 2 cos θ−1=0 Or 2 cos θ+5=0⇒cos θ=12 Or cos θ=−52[This is not possible as−1<cos θ<1]⇒cos θ=cosπ3⇒ θ=2nπ±π3,n∈z(v) tan2 x+(1−√3)tan x−√3=0We have,tan2 x+(1−√3)tan x−√3=0⇒tan2 x+tan x−√3 tan x−√3=0⇒tan x(tan x+1)−√3(tan x+1)=0⇒(tan x−√3)(tan x+1)=0⇒eithertan x=√3 Or tan x=−1⇒tan x=tan π3Or tan x=−tan π4⇒x=nπ+ π3,n∈z Or x=mπ−π4,m∈z∴x=nπ+ π3Or mπ−π4,n,m∈z(vi) 3 cos2θ−2√3 sinθ cos θ−3 sin2θ=0√3 cos2 θ−sin θ cos θ−√3 sin2 θ=0 (Divideni by√3)√3 cos2 θ+2 sin θ cos θ−3 sin θ cos θ−√3 sin2 θ=0cos θ(√3 cos θ+sin θ)−√3 sin θ(√3 cos θ+sin θ)=0(√3 cos θ+sin θ)(cos θ−√3 sin θ)=0√3 cos θ+sin θ=0Or cos θ−√3 sin θ=0tan θ=−√3=−tanπ3Or tan θ= 1√3=tanπ6 θ=nπ−π3 Or θ=mπ+π6n,m∈z(vii) cos 4θ=cos 2θWe have,cos 4θ=cos 2θ⇒cos 4θ−cos 2θ=0⇒2 sin θ.sin 3θ=0⇒eithersin θ=0 Or sin 3θ=0⇒θ=nπ,n∈z Or 3θ=mπ, m∈zThus,θ=nπ, Or m π3,n ,m ∈z
