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2Cos^(2)theta+Cos^(2)2 theta-2Cos^(2)theta*Cos2 theta=1 |
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Answer» As Sasha pointed out, one can simply add cos2θ to both SIDES and solve for the quadratic EQUATION. Here is what occurs: 2cos2θ−cosθ−1+cos2θ=sin2θ+cos2θ 3cos2θ−cosθ−1=1 3cos2θ−cosθ−2=0 Let w=cosθ. Then we get: 3w2−w−2=0 3w2−3w+2w−2=0 3w(w−1)+2(w−1)=0 (3w+2)(w−1)=0 So w=−23 or w=1, which MEANS cosθ=−23 or cosθ=1. This gives us the answers θ=2kπ,k∈Z and θ=±cos−1(−23)+2kπ,k∈Z |
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