To solve the equation cos²(2x) + 2 cos(x) + 1 = 0, we begin by simplifying and analyzing the components of the expression.
First, we rearrange the equation:
cos²(2x) = -2 cos(x) – 1
Since the left side, cos²(2x), must always be non-negative (as the cosine function squared cannot be less than zero), the right side of the equation must also be non-negative. Therefore, we need to analyze when -2 cos(x) – 1 is non-negative:
-2 cos(x) – 1 ≥ 0
Solving this inequality:
-2 cos(x) ≥ 1
Thus, we find:
cos(x) ≤ - rac{1}{2}
The cosine function is less than or equal to -1/2 in the second and third quadrants. Specifically, within the interval from 0 to 2π, this occurs at:
- x = 2π/3 + 2kπ (for the second quadrant)
- x = 4π/3 + 2kπ (for the third quadrant)
where k is any integer. Now, substituting these values back into the original equation does not yield additional restrictions on the possible values of 2x.
Next, let’s find the solutions for 2x. From x = 2π/3 + 2kπ and x = 4π/3 + 2kπ, we have:
- 2x = 4π/3 + 4kπ
- 2x = 8π/3 + 4kπ
Finally, dividing the entire equation by 2 gives:
- x = 2π/3 + 2kπ (from the first case)
- x = 4π/3 + 2kπ (from the second case)
In conclusion, the solutions to the equation cos²(2x) + 2 cos(x) + 1 = 0 are:
- x = 2π/3 + 2kπ
- x = 4π/3 + 2kπ
where k is any integer.