Question Number 102783 by bramlex last updated on 11/Jul/20

2(√3)−1 = 6sin (2θ−60^o )−2sin  (2θ−30^o ) ⇒θ=?

Answered by 1549442205 last updated on 11/Jul/20

⇔6sin2θ×(1/2)−6cos2θ×((√3)/2)−(2sin2θ×((√3)/2)−2cos2θ×(1/2))=2(√3)−1  ⇔(3−(√3))sin2θ−(3(√3)−1)cos2θ=2(√3)−1(1)  Put x=tanθ.We have  (1)⇔(3−(√3))((2x)/(1+x^2 ))−(3(√3)−1)((1−x^2 )/(1+x^2 ))=2(√3)−1(2)  2(3−(√3))x−(3(√3)−1)(1−x^2 )=(2(√3)−1)(1+x^2 )  (6−2(√3))x−3(√3)+1+(3(√3)−1)x^2 −2(√3)+1−(2(√3)−1)x^2 =0  (√3)x^2 +2(3−(√3))x−(5(√3)−2)=0  tan𝛉=x_1 =1.361080303⇔𝛉≈53°41′41  tan𝛉=x_2 =−2.825181918⇔𝛉≈−70°30′30

Answered by bemath last updated on 11/Jul/20

Commented byfloor(10²Eta[1]) last updated on 11/Jul/20

why the eq. has solution if that  inequality happens?