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

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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.` **