How do you find the solution to $2 cos θ - 1 = 0$ $2costheta-1=0$ if $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2017-04-27T20:33:50

$S = {60^{\circ}, 300^{\circ}}$ $S={60^@,300^@}$

$2 cos θ - 1 = 0$ $2 cos theta-1=0$

$2 cos θ = 1$ $2cos theta = 1$

$cos θ = \frac{1}{2}$ $cos theta = 1/2$

$θ = {cos}^{- 1} (\frac{1}{2})$ $theta = cos^-1 (1/2)$

$θ = \pm 60^{\circ} + 360^{\circ} n$ $theta=+- 60^@ + 360^@ n$ , where n are integers

$n = 0, θ = 60^{\circ}, - 60^{\circ}$ $n=0, theta = 60^@, -60^@$

$n = 1, θ = 420^{\circ}, 300^{\circ}$ $n=1, theta = 420^@, 300^@$

$S = {60^{\circ}, 300^{\circ}}$ $S={60^@,300^@}$

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