How do you solve the equation on the interval [0,2pi) for cos^2 theta - cos theta - 2 = 0 cos2θ−cosθ−2=0?

1 Answer
Nghi N.
Jul 22, 2016

t = pit=π

Explanation:

f(t) = cos^2 t - cos t - 2 = 0
This is a quadratic equation for cos t.
Since a - b + c = 0, use shortcut. Two real roots: cos t = - 1 and
cos t = -c/a = 2
a. cos t = - 1 --> t = pit=π
b. cos t = 2 (rejected since > 1)
Answer for (0, 2pi)(0,2π)
t = pit=π
Check
t = pit=π --> cos ^2 t = 1cos2t=1 --> cos t = -1.
cos^2 t - cos t - 2 = 1 - (-1) - 2 = 0cos2t−cost−2=1−(−1)−2=0. OK

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