I'm trying to find all solutions to the equation 5cos(x+3)=1 from 0 to 2pi and can't figure out why my answer isn't matching the back of the...

There is no mistake in your answer but something it's
missing.

To solve an elementary trigonomteric equation
involving cosine function, you'll have to do the following
steps:

cos x = a

x = `+-`
arccos (a) + 2k*`pi` , where k`in` Z, the set of integer
numbers

Therefore, we'll solve the equation from the
point:

cos(x+3) = 1/5

x + 3 =
`+-` arccos(1/5) + 2k*`pi`

x = `+-` arccos(1/5)  +
2k`pi`

We'll consider k =0 => 2k`pi` =
0

arccos (1/5) = arccos(0.2) = 0.064
`pi`

x = 0.064 `pi` - 3

If
we'll consider pi = 3.14, we'll get:

x = 0.20096 -
3

x = -2.79 approx.

x = -
0.064 `pi` - 3

x = -3.20
approx.

We'll consider k =1 => 2k`pi` =
2`pi`

x = 0.064`pi` + 2`pi` -
3

x = 3.48 approx.

x = -0.064
`pi` + 2`pi` - 3

x = 3.07
approx

Since the angle is included in the interval [0,2`pi`
], then k cannot be larger than 1.

Therefore,
the solutions of the equation, in the interval [0,2`pi ` ], for `pi` = 3.14, are: {-3.20
; -2.79 ; 3.07 ; 3.48}.