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

1 answer:

4 0

Hello,

cos 2t=2cos²t-1

2cos 2t -cos t-1=
==>2(2cos² t -1)-cos t-1=0
==>4cos² t -cos t -3=0
Δ=(1+4*4*3=49=7²
cos t= (1+7)/8 or cos t=(1-7)/8
cos t =1 or cos t=-3/4

t=0° or t=138.590377...°

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