# 2cos2θ−cosθ−1=0 for −180°≤θ≤180° I don't get how it is solved

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## Answers

Hello,cos 2t=2cos²t-12cos 2t -cos t-1===>2(2cos² t -1)-cos t-1=0==>4cos² t -cos t -3=0Δ=(1+4*4*3=49=7²cos t= (1+7)/8 or cos t=(1-7)/8cos t =1 or cos t=-3/4t=0° or t=138.590377...°

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