SOLUTION: solve for solutions over the interval (0 deg, 360 deg) 4 cos^2 theta + 4 cos theta = 1 i believe the quadratic formula applies but am stuck.

Question 955454: solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta = 1
i believe the quadratic formula applies but am stuck.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta - 1 = 0
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cos(t) = [-4 +- sqrt(16-4*4*-1)]/8 = [-4+-4sqrt(2)]/8
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cos(t) = [-4 + 4sqrt(2)]/8 = [-1+sqrt(2)]/2 = 0.2071
or
cos(t) [-1-sqrt(2)]/2 = -1.2071 (extraneous)
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t = arccos(0.2071) = 78.04 degrees or 281.96 degrees
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Cheers,
Stan H.
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