Question 993622
Please help me out with this problem! Solve for all solutions of t on the interval [0,2pi)

2-2sin(t)=2sqrt(3)cos(t)
{{{1 - sin(t) = sqrt(3)*cos(t)}}}
Square both sides.
{{{1 - 2sin(t) + sin^2(t) = 3cos^2(t)}}}
Sub 1 - sin^2 for cos^2
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{{{1 - 2sin(t) + sin^2(t) = 3 - 3sin^2(t)}}}
{{{4sin^2(t) - 2sin(t) - 2 = 0}}}
{{{2sin^2(t) - sin(t) - 1 = 0}}}
(2sin + 1)*(sin - 1) = 0
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sin(t) = 1
t = pi/2
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sin(t) = -1/2
t = 7pi/6, 11pi/6