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Mathematics, 10.10.2019 03:30 AaronEarlMerringer

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

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

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