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Question Number 47150 by somil last updated on 05/Nov/18

Commented by maxmathsup by imad last updated on 05/Nov/18

we see that cosθ must be ≥0⇒θ ∈[−(π/2),(π/2)][2π]  changement =x =sinθ give  3(√(1−x^2 ))=2x^2  ⇒9(1−x^2 )=4x^4  ⇒4x^4 +9x^2  −9=0  x^2 =u ⇒4u^2  +9u−9=0 ⇒Δ=9^2 −4.4.(−9) =9(9 +16)=9.25 ⇒(√Δ)=15  u_1 =((−9+15)/8) =(6/8) =(3/4) and u_2 =((−9−15)/8) =−3<0(to eliminate) ⇒  x^2 =(3/4) ⇒x=+^− ((√3)/2)     case 1   x=((√3)/2) ⇒sinθ =sin((π/3)) ⇒ θ =(π/3) +2kπ or θ =((2π)/3) +2kπ  −(π/2)≤(π/3)+2kπ≤(π/2) ⇒−(1/2)≤(1/3) +2k≤(1/2) ⇒−(5/3) ≤2k≤(1/6) ⇒−(5/6)≤k≤(1/(12))  ⇒k=0 ⇒θ =(π/3)  −(π/2)≤((2π)/3) +2kπ≤(π/2) ⇒−(1/2) ≤(2/3) +2k≤(1/2) ⇒((−7)/3) ≤2k≤−(1/6) ⇒  −(7/6)≤k≤−(1/(12)) ⇒k=−1 ⇒x=((2π)/3) −2π =−((4π)/3)  and cos(((4π)/3))<0 (toeliminate)  case 2  x =−((√3)/2) ⇒sinθ =sin(−(π/3)) ⇒θ =−(π/3) +2kπ or θ=((4π)/3) +2kπ  we follow the same manner to find x ...

weseethatcosθmustbe0θ[π2,π2][2π]changement=x=sinθgive31x2=2x29(1x2)=4x44x4+9x29=0x2=u4u2+9u9=0Δ=924.4.(9)=9(9+16)=9.25Δ=15u1=9+158=68=34andu2=9158=3<0(toeliminate)x2=34x=+32case1x=32sinθ=sin(π3)θ=π3+2kπorθ=2π3+2kππ2π3+2kππ21213+2k12532k1656k112k=0θ=π3π22π3+2kππ21223+2k12732k1676k112k=1x=2π32π=4π3andcos(4π3)<0(toeliminate)case2x=32sinθ=sin(π3)θ=π3+2kπorθ=4π3+2kπwefollowthesamemannertofindx...

Commented by maxmathsup by imad last updated on 05/Nov/18

another method but easy  (e) ⇔ 3cosθ =2(1−cos^2 θ) ⇒  3cosθ−2 +2cos^2 θ =0 ⇒2 cos^2 θ +3cosθ −2 =0  with condition cosθ≥0 ⇒  θ ∈[−(π/2),(π/2)]  changement cosθ =x drive to 2 x^2  +3x−2 =0  Δ =9−4.2(−2) =9+16 =25 ⇒x_1 =((−3+5)/4) =(1/2) and x_2 =((−3−5)/4) <0(to eliminate)  after we solve the equation cosθ =(1/2) ....

anothermethodbuteasy(e)3cosθ=2(1cos2θ)3cosθ2+2cos2θ=02cos2θ+3cosθ2=0withconditioncosθ0θ[π2,π2]changementcosθ=xdriveto2x2+3x2=0Δ=94.2(2)=9+16=25x1=3+54=12andx2=354<0(toeliminate)afterwesolvetheequationcosθ=12....

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

3cosθ=2(1−cos^2 θ)  3x=2(1−x^2 )  2x^2 +3x−2=0  2x^2 +4x−x−2=0  2x(x+2)−1(x+2)=0  (x+2)(2x−1)=0  (cosθ+2)(2cosθ−1)=0  cosθ can not be equals to −2  2cosθ−1=0  cosθ=(1/2)=cos((π/3))  θ=2nπ±(π/3)

3cosθ=2(1cos2θ)3x=2(1x2)2x2+3x2=02x2+4xx2=02x(x+2)1(x+2)=0(x+2)(2x1)=0(cosθ+2)(2cosθ1)=0cosθcannotbeequalsto22cosθ1=0cosθ=12=cos(π3)θ=2nπ±π3

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