How do you solve 2cos^2(x-5)-3cos(x-5)+1=02cos2(x−5)−3cos(x−5)+1=0?

1 Answer
Nghi N.
Jul 26, 2015

Solve y = 2cos^2 (x - 5) - 3 cos (x - 5) + 1 = 0

Explanation:

Call cos (x - 5 deg) = t
2t^2 - 3t + 1 = 0
Since a + b + c = 0, use shortcut. 2 real roots: t = 1 and t = 1/2
t = cos (x - 5) = 1 = cos 0 --> x = 5x=5 deg
t = cos (x - 5) = 1/2 = +- cos 60 cos(x−5)=12=±cos60-->

a. x - 5 = 60 --> x = 65x=65 deg
b. x - 5 = -60 --> x = - 55x=−55 deg

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