Question

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

Answer 1

Solve 2cos^2 (x/5) - 3cos (x/5) + 1 = 0
Ans: 0, `10pi.` `+- (5pi)/3`

Explanation:

Call `cos (x/5) = t` and solve quadratic equation:

`2t^2 - 3t + 1 = 0`

Since a + b + c = 0, two real roots are: 1 and `c/a = 1/2.`
Next, solve:
a. `t = cos (x/5) = 1`-->

`x/5 = 0`, --> `x = 0`

and `x/5 = 2pi` --> `x = 10pi`

b.`t = cos (x/5) = 1/2` --> `x/5 = +- pi/3` --> `x = +- (5pi)/3`