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)[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."cos2(3x)=sin2(3x)
Question
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)[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."cos2(3x)=sin2(3x)
Solution
To solve the equation cos²(3x) = sin²(3x), we can use the Pythagorean identity sin²θ + cos²θ = 1. This can be rearranged to sin²θ = 1 - cos²θ or cos²θ = 1 - sin²θ.
In this case, we can substitute sin²(3x) with 1 - cos²(3x) in the equation.
So, the equation becomes cos²(3x) = 1 - cos²(3x).
Solving this equation, we get:
2cos²(3x) = 1
cos²(3x) = 1/2
Taking the square root of both sides, we get:
cos(3x) = ± √(1/2)
cos(3x) = ± √2/2
Now, we find the values of x that satisfy the equation.
For cos(3x) = √2/2, the solutions in the interval [0, 2π) are π/4 and 7π/4.
For cos(3x) = -√2/2, the solutions in the interval [0, 2π) are 3π/4 and 5π/4.
However, we need to divide these solutions by 3 (because of the 3x in the cos function), and also consider the periodicity of the cosine function, which is 2π.
So, the solutions are:
For cos(3x) = √2/2:
x = π/12 + 2nπ/3, 7π/12 + 2nπ/3
For cos(3x) = -√2/2:
x = π/4 + 2nπ/3, 5π/12 + 2nπ/3
where n is any integer.
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