cos2x(2cosx+1)=0

Solve for x.

Please.

(2cos^2x-1)(2cosx+1)=0

^
[Double angle identity]

Now I'm stuck.

hey ted did you ever figure that out because if so tell me tell me now plz i need this very much

To solve the equation cos(2x)(2cos(x) + 1) = 0 for x, we need to find the values of x that satisfy the equation.

First, let's start by solving the equation cos(2x) = 0.

The cosine function is equal to zero at two points within each period: at π/2 and 3π/2.

To find the values of x for cos(2x) = 0, we can set 2x equal to these values:

2x = π/2 + kπ (where k is an integer)
2x = 3π/2 + kπ (where k is an integer)

Solving each equation for x:

x = (π/2 + kπ) / 2 (where k is an integer)
x = (3π/2 + kπ) / 2 (where k is an integer)

Now let's solve the second part of the equation: 2cos(x) + 1 = 0.

Subtracting 1 from both sides:
2cos(x) = -1

Dividing both sides by 2:
cos(x) = -1/2

The cosine function is equal to -1/2 at two points within each period: π/3 and 5π/3.

To find the values of x for cos(x) = -1/2, we can set x equal to these values:

x = π/3 + kπ (where k is an integer)
x = 5π/3 + kπ (where k is an integer)

So, the solutions for x in the equation cos(2x)(2cos(x) + 1) = 0 are the values of x that satisfy either cos(2x) = 0 or cos(x) = -1/2.

That gives us the following possible solutions:

x = (π/2 + kπ) / 2 (where k is an integer)
x = (3π/2 + kπ) / 2 (where k is an integer)
x = π/3 + kπ (where k is an integer)
x = 5π/3 + kπ (where k is an integer)

Therefore, to find the specific values of x that satisfy the equation, substitute different integer values for k into each equation.