How do you solve $3 - 3 cos (θ) = 2 {sin}^{2} (θ)$ $3- 3 Cos (theta) = 2 Sin^2 (theta)$ ?

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How do you solve $3 - 3 cos (θ) = 2 {sin}^{2} (θ)$ $3- 3 Cos (theta) = 2 Sin^2 (theta)$ ?

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Answer 1 · 2016-04-13T02:54:28

In $[0, 2 π], θ = 0, \frac{π}{3}, 5 \frac{π}{3} and 2 π$ $[0, 2pi], theta = 0, pi/3, 5pi/3 and 2pi$ .
General solution: $θ = 2 k π and θ = 2 k π \pm \frac{π}{3}$ $theta=2kpi and theta= 2kpi+-pi/3$ , k=1,2,3,....

Explanation:

Using ${sin}^{2} θ = 1 - {cos}^{2} θ and c = cos θ$ $sin^2theta=1-cos^2theta and c=costheta$ ,,
$3 - 3 c = 2 (1 - c^{2})$ $3-3c=2(1-c^2)$
$2 c^{2} - 3 c + 1 = 0$ $2c^2-3c+1=0$
$c = cos θ = 1, \frac{1}{2}$ $c=cos theta=1, 1/2$ ..

In $[0, 2 π] : a r c cos 1 = 0 and 2 π$ $[0, 2pi]: arc cos 1=0 and 2pi$ .
$a r c cos (\frac{1}{2}) = \frac{π}{3} and \frac{5 π}{3}$ $arc cos(1/2)=pi/3 and (5pi)/3$ ,
using $cos (2 π \pm θ) = cos θ$ $cos(2pi+-theta)=costheta$

General solution is given by $cos (2 k π \pm θ) = cos θ$ $cos(2kpi+-theta)=costheta$ , k=1,2,3....

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How do you solve $3 - 3 cos (θ) = 2 {sin}^{2} (θ)$ $3- 3 Cos (theta) = 2 Sin^2 (theta)$ ?

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How do you solve $3 - 3 cos (θ) = 2 {sin}^{2} (θ)$ $3- 3 Cos (theta) = 2 Sin^2 (theta)$ ?