Solve using factoring

Conorgriffin asked Mar 20, 2023

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Conorgriffin asked Mar 20, 2023

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10 votes

Answer:

${ \tt{4 { \cos }^(2) x - 1 = 0}} \\ { \tt{ {(2 \cos x) }^(2) - {1}^(2) = 0 }} \\ { \tt{(2 \cos x - 1)(2 \cos x + 1) = 0}}$

» Either 2cos x -1 or 2cos x +1 is zero

${ \tt{2 \cos x = \pm 1 }} \\ { \tt{ \cos x = \pm (1)/(2) }} \\ \\ { \tt{x = 60 \degree \: and \: 120 \degree}}$

TZubiri answered Mar 24, 2023

3.0k points

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