how do you solve 2sin(theta) = cos(theta/3) ?
2sin(θ)=cos(θ3) we set θ=3α2sin(3α)=cos(α)
Formula:2sin(3α)=cos(α)(1)sin(3α)=sin(α+2α)=sin(α)cos(2α)+cos(α)sin(2α)cos2α=1−2sin2(α)sin(3α)=sin(α)(1−2sin2(α))+2sin(α)cos2(α)sin2α=2sin(α)cos(α)sin(3α)=sin(α)(1−2sin2(α))+2sin(α)(1−sin2(α))cos2α=1−sin2(α)⋯sin(3α)=sin(α)[(3−4sin2(α)]2sin(α)[(3−4sin2(α)]=cos(α)2[(3−4sin2(α)]=cot(α)⋯8sin2(α)=6−cot(α)1sin2(α)=1+cot2(α)8=[(6−cot(α)][1+cot2(α)]⋯cot3(α)−6cot2(α)+cot(α)+2=0α=θ3cot3(θ3)−6cot2(θ3)+cot(θ3)+2=0
cot3(θ3)−6cot2(θ3)+cot(θ3)+2=0 substitute: u=cot(θ3)=1tan(θ3)θ=3arctan( 1u )±3π⋅kk=0,1,2,3⋯ u3−6u2+u+2=0 u1=5.766435484θ1=3arctan(1u1)θ1=0.515128919±3π⋅ku2=−0.483611621θ2=3arctan(1u2)θ2=−3.361035503±3π⋅ku3=0.717176136θ3=3arctan(1u3)θ3=2.845906583±3π⋅k
Compare with wolframalpha.com http://www.wolframalpha.com/input/?i=2*sin%28x%29%3Dcos%28x%2F3%29 :
we have seen: cot3(θ3)−6cot2(θ3)+cot(θ3)+2=0
we set cot(θ3)=1−tan2(θ6)2tan(θ6)and use x=tan(θ6)then cot(θ3)=1−x22xwe substitute(1−x22x)3−6(1−x22x)2+(1−x22x)+2=0(1−x2)38x3−6(1−x2)24x2+(1−x2)2x+2=0|⋅8x3(1−x2)3−12x(1−x2)2+4x2(1−x2)+16x3=0⋯1−3x2+3x4−x6−12x+24x3−12x5+4x2−4x4+16x3=0finallyx6+12x5+x4−40x3−x2+12x−1=0wolframalpha.com solution:x6+12x5+x4−40x3−x2+12x−1=0withθ6=arctan(x)±π⋅kθ=6[arctan(x)±π⋅k]θ=6arctan(x)±6π⋅k