4cos(2x)+2cos(x)+3=0

substitute:

cos(2x)=cos²(x)-sin²(x)
=2cos²(x)-1
to get
8cos²(x)+2cos(x)-1=0
let y=cos(x),
8y²+2y-1=0
(4y-1)(2y+1)=0
y=1/4, y=-1/2
cos(x)=1/4, cos(x)=-1/2
therefore
x=2kπ±acos(1/4), or
x=2kπ±acos(-1/2)
You may want to express the solutions in a more familiar form.

To solve the equation 4cos(2x) + 2cos(x) + 3 = 0, we can use the trigonometric identity for double angle: cos(2x) = 2cos²(x) - 1.

Replace cos(2x) in the equation with 2cos²(x) - 1:

4(2cos²(x) - 1) + 2cos(x) + 3 = 0

Simplify the equation:

8cos²(x) - 4 + 2cos(x) + 3 = 0

Rearrange the terms:

8cos²(x) + 2cos(x) - 1 = 0

Now, let's substitute u = cos(x) in the equation:

8u² + 2u - 1 = 0

Now we have a quadratic equation in terms of u. We can solve this by factoring, completing the square, or using the quadratic formula.

Unfortunately, in this case, the equation cannot be factored easily. Let's use the quadratic formula:

u = (-b ± √(b² - 4ac)) / (2a)

Here, a = 8, b = 2, and c = -1. Plugging these values into the quadratic formula:

u = (-2 ± √(2² - 4 * 8 * -1)) / (2 * 8)

Simplify:

u = (-2 ± √(4 + 32)) / 16

u = (-2 ± √36) / 16

u = (-2 ± 6) / 16

Now we have two possible values for u:

u₁ = (6 - 2) / 16 = 4 / 16 = 1/4

u₂ = (-6 - 2) / 16 = -8 / 16 = -1/2

These values represent possible values of cos(x). To find the values of x, we can take the inverse cosine (also known as arccos) of these solutions:

cos(x) = 1/4
x = arccos(1/4)

cos(x) = -1/2
x = arccos(-1/2)

These calculations can be done using a scientific calculator or trigonometric tables. Remember that there are infinitely many solutions since the cosine function is periodic.