You need to write the equation in terms of `cos x` , hence, you need to use the double angle identity and half angle identity, such that:

`cos 2x = 2cos^2 x - 1`

`cos^2(x/2) = (1 + cos x)/2`

Replacing `2cos^2 x - 1` for `cos 2x` and `(1 + cos x)/2` for `cos^2(x/2)` in equation, yields:

`2cos^2 x - 1 - 3cos x = 4*(1 + cos x)/2`

`2cos^2 x - 1 - 3cos x = 2(1 + cos x)`

`2cos^2 x - 1 - 3cos x = 2 + 2 cos x`

You need to move the terms to the left side, such that:

`2cos^2 x - 3cos x - 2cos x - 1 - 2 = 0`

`2cos^2 x - 5cos x - 3 = 0`

You should come up with the substitution `cos x = y` , such that:

`2y^2 - 5y - 3 = 0`

Using quadratic formula, yields:

`y_(1,2) = (5+-sqrt(25 + 24))/4 => y_(1,2) = (5+-sqrt49)/4`

`y_(1,2) = (5+-7)/4 => y_1 = 3 ; y_2 = -1/2`

You need to solve for x the following equations, such that:

`cos x = 3` invalid since `cos x ` cannot be larger than `1`

`cos x = -1/2 => x = +-cos^(-1)(-1/2) + 2*n*pi`

`x = pi +- pi/3 + 2*n*pi => {(x = (4pi)/3 + 2n*pi),(x = (2pi)/3 + 2n*pi):}`

**Hence, evaluating the general solutions to the given equation, yields **`x = (2pi)/3 + 2n*pi, x = (4pi)/3 + 2n*pi.`

**Further Reading**

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