sin 2x + 2 cos ^2 x = 0

You should
substitute `2 sin x cos x`  for `sin 2x`   such that:

`2 sin x cos x + 2
cos^2 x = 0`

You need to factor out `2 cos x`  such that:

`2 cos x(sin x + cos x) = 0`

You need to solve the equation `2 cos x
= 0`  such that:

`2 cos x = 0 => cos x = 0 => x = +-arccos 0 +
2npi`

`x = +-pi/2 + 2n pi`

You need to solve `sin x + cos
x = 0,`  hence you need to divide by `cos x`  such that:

`sin x/cos x +1 = 0
=> tan x + 1 = 0 => tan x = -1`

`x = arctan(-1) + npi => x = -
arctan 1 + npi`

`x = -pi/4 + npi`

Hence,
evaluating the general solutions to the given equation yields `x = +-pi/2 + 2n pi ` and `x =
-pi/4 + npi.`