Question

How do you solve `2cos(3x + pi/4) =1`?

Answer 1

The general solution is `(24n+1)pi/36 and (24n-7)pi/36, n = 0, 1, 2, 3,...`,
`n=0: pi/36 in (0, pi/2) and -7/36pi in (-pi/2, 0).`The solution `in (0, pi)` are `pi/36 and (25/36)pi`.

Explanation:

`cos(3x+pi/4)=1/2=cos(+-pi/3,)` and, for the general solution,

`cos(3x+pi/4)=1/2=cos(2npi+-pi/3), n=0, 1, 2, ..`

So, the general solution is

`3x+pi/4=2npi+-pi/3, n=0, 1, 2, ..`. Solving,

`x=(24n+1)pi/36 and x=(24n-7)pi/36, n = 0, 1, 2, 3,...`,

The solution `in (0, pi)` are `pi/36 and (25/36)pi`.

The second comes from n=1 of `x=(24n+1)pi/36`.