Question

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

Answer 1

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 = 5` deg
t = `cos (x - 5) = 1/2 = +- cos 60`-->

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