You should use the following trigonometric identities to solve the equation `1 + sin 2x = 3 cos 2x` such that:

sin 2x = (2 tan ((2x)/2))/(1 + tan^2((2x)/2))

`sin 2x = (2 tan x)/(1 + tan^2 x)`

`cos2x = (1- tan^2 x)/(1 + tan^2 x)`

`1 + (2 tan...

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You should use the following trigonometric identities to solve the equation `1 + sin 2x = 3 cos 2x` such that:

sin 2x = (2 tan ((2x)/2))/(1 + tan^2((2x)/2))

`sin 2x = (2 tan x)/(1 + tan^2 x)`

`cos2x = (1- tan^2 x)/(1 + tan^2 x)`

`1 + (2 tan x)/(1 + tan^2 x) = 3(1 - tan^2 x)/(1 + tan^2 x)`

Moving the terms to one side yields:

`1 + (2 tan x)/(1 + tan^2 x)- 3(1 - tan^2 x)/(1 + tan^2 x) = 0`

Bringing the terms to a common denominator yields:

`(1 + tan^2 x + 2tan x - 3 + 3tan^2 x)/(1 + tan^2 x) = 0`

Since `1 + tan^2 x != 0` , hence `4tan^2 x + 2tan x - 2 = 0` such that:

`4tan^2 x + 2tan x - 2 = 0 => 2tan^2 x + tan x - 1 = 0`

You should come up with the following substitution such that:

`tan x = t`

`2t^2 + t - 1 = 0`

Using quadratic formula yields:

`t_(1,2) = (-1 +- sqrt(1 + 8))/2 => t_(1,2) = (-1 +- 3)/2`

`t_1 = 1 ; t_2 = -2`

You need to solve the equations `tan x = t_1` and `tan x = t_2` such that:

`tan x = 1 => x = pi/4 and x = pi + pi/4 => x = 5pi/4`

`tan x = -2 => x = arctan(-2) => x = -arctan 2`

**Hence, evaluating the solutions to the equation yields `x = pi/4, x = 5pi/4` and `x = -arctan 2` .**