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2cos^2(t) + 2cos(t) - 1 = 0

This doesn't factor, so we have to use the quadratic formula.

cos(t) = [-2 +/- sqrt(4 - 4(2)(-1))]/[4]
cos(t) = [-2 +/- sqrt(12)]/4
cos(t) = [-2 +/- 2sqrt(3)]/4
cos(t) = [-1 +/- sqrt(3)]/2

Thus, we get two equations:

cos(t) = [-1 + sqrt(3)]/2
cos(t) = [-1 - sqrt(3)]/2

These values are not known on the unit circle. Therefore, assuming your answers are restricted to [0, 2pi), we have

t = { cos^(-1) {[-1 + sqrt(3)]/2} , pi - cos^(-1) {[-1 + sqrt(3)]/2} }
t = { cos^(-1) {[-1 - sqrt(3)]/2} , pi - cos^(-1) {[-1 - sqrt(3)]/2} }

Four solutions (despite how ugly they look).

2007-01-08 13:22:41 · answer #1 · answered by Puggy 7 · 0 0

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