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.`
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