Question

How do you solve `cos^2(3theta)-5cos(3theta)+4=0`?

Answer 1

Please see the explanation.

Explanation:

Please notice that the quadratic factors into:

`(cos(3theta) - 1)(cos(3theta) - 4) = 0`

Which implies

`cos(3theta) = 1 and cos(3theta) = 4`

However, `cos(3theta) = 4` must be discarded, because it exceeds the range of the cosine function.

This leaves:

`cos(3theta) = 1`

To make the cosine function disappear, take the inverse cosine of both sides:

`3theta = cos^-1(1)`

The inverse cosine of 1 is zero:

`3theta = 0`

But it repeats every integer multiple of `2pi` from `-oo` to `oo`:

`3theta = 0 + 2npi; n = ...,-3, -2, -1, 0, 1, 2, 3,...`

Divide by 3:

`theta = 2/3npi; n = ...,-3, -2, -1, 0, 1, 2, 3,...`

Answer 2

`t = (2kpi)/3`

Explanation:

Call 3t = T, and solve the quadratic equation for cos T:
`cos^2 T - 5cos T + 4 = 0`
`D = d^2 = 25 - 16 = 9`--> `d = +- 3`
There are 2 real roots:
`cos T = -b/(2a) +- d/(2a) = 5/2 +- 3/2`
cos T1 = 4 (rejected as > 1)
`cos T2 = (5 - 3)/2 = 1`
cos T = cos 3t = 1
Trig unit circle -->

a/ `3t = 0 + 2kpi` -->
`t = (2kpi)/3`
b/ `3t = 2pi + 2kpi`-->
`t = (2pi)/3 + (2kpi)/3`
General answer: `t = (2kpi)/3`