How do you solve 1=cot^2theta+csctheta1=cot2θ+cscθ for 0<=theta<=2pi0θ2π?

1 Answer
Sep 20, 2016

Apply the identities cottheta = costheta/sinthetacotθ=cosθsinθ and csctheta = 1/sinthetacscθ=1sinθ:

cos^2theta/sin^2theta + 1/sintheta = 1cos2θsin2θ+1sinθ=1

cos^2theta/sin^2theta + sintheta/sin^2theta = 1cos2θsin2θ+sinθsin2θ=1

(cos^2theta + sin theta)/sin^2theta = 1cos2θ+sinθsin2θ=1

cos^2theta + sin theta = sin^2thetacos2θ+sinθ=sin2θ

Apply the identity sin^2theta + cos^2theta = 1 ->cos^2theta = 1 - sin^2thetasin2θ+cos2θ=1cos2θ=1sin2θ

1 - sin^2theta + sin theta - sin^2theta = 01sin2θ+sinθsin2θ=0

-2sin^2theta + sin theta + 1 = 02sin2θ+sinθ+1=0

-2sin^2theta + 2sintheta - sin theta + 1 = 02sin2θ+2sinθsinθ+1=0

-2sintheta(sin theta - 1) - 1(sin theta - 1) = 02sinθ(sinθ1)1(sinθ1)=0

(-2sintheta - 1)(sin theta - 1) = 0(2sinθ1)(sinθ1)=0

sintheta = -1/2 and sin theta = 1sinθ=12andsinθ=1

theta = (7pi)/6, (11pi)/6 and pi/2θ=7π6,11π6andπ2

However, since pi/2π2 renders the equation undefined, that solution is extraneous. Hence, our solution set is {(7pi)/6, (11pi)/6}{7π6,11π6}.

Hopefully this helps!