The trigonometric equation 2cos^2(x) + 2 = 4 has solutions x = 0 and x = π in the interval [0, 2π). To include all possible solutions, we express them as x = 2nπ and x = π + 2nπ with n as any integer.
To solve the trigonometric equation 2cos^2(x) + 2 = 4, we start by simplifying the equation:
- Subtract 2 from both sides to get 2cos^2(x) = 2.
- Divide both sides by 2 to get cos^2(x) = 1.
- Take the square root of both sides, considering both positive and negative roots, to get cos(x) = ±1.
- Find angles x in the interval [0, 2π) where the cosine is ±1. For cos(x) = 1, x = 0. For cos(x) = -1, x = π.
This equation has two solutions in the interval [0, 2π): x = 0 and x = π (which are 0 and 3.1416 when rounded to four decimal places). To express the infinite set of solutions that occur at every period of cos(x), we add 2nπ to each solution, where n is any integer, giving the final solutions as x = 2nπ and x = π + 2nπ.
the complete Question is given below:
Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0, 2π)
and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."
2cos^2 (x) + 2 = 4
Enter your answer in radians, as an exact answer when possible. Multiple answers should be separated by commas.