How do I find all solutions in the interval [0,2Pi):
sin2x=cos2x
answer choices are:
a.)Pi/8
b.) Pi/8, 5Pi/8, 9Pi/8, 13Pi/8
The solution is tan 2x = 1
2x = arctan 1
2x = pi/4 or (5/4) pi or (9/4) pi or (13 pi/8). You can ignore higher values because of the 2 pi limit on x.
Solve for x
To find all the solutions to the equation sin(2x) = cos(2x) in the interval [0, 2π), we can use trigonometric identities to simplify the equation.
1. Start by using the double angle formulas for sine and cosine:
sin(2x) = cos(2x)
2sin(x)cos(x) = cos^2(x) - sin^2(x)
2. Rearrange the equation to one side:
2sin(x)cos(x) - (cos^2(x) - sin^2(x)) = 0
3. Apply the Pythagorean identity sin^2(x) + cos^2(x) = 1:
2sin(x)cos(x) - (1 - 2sin^2(x)) = 0
4. Distribute and simplify the equation:
2sin(x)cos(x) - 1 + 2sin^2(x) = 0
5. Combine like terms and rearrange:
2sin^2(x) + 2sin(x)cos(x) - 1 = 0
6. Factor out a common factor of 2:
2(sin^2(x) + sin(x)cos(x)) - 1 = 0
7. Use the identity sin(2θ) = 2sin(θ)cos(θ) to further simplify:
sin(x) + sin(2x) - 1 = 0
Now we have a single trigonometric function in terms of x. We can solve this equation by finding the values of x that make sin(x) + sin(2x) equal to 1.
To find the solutions in the interval [0, 2π), you can use a numerical method like graphing or Newton's method, or you can use a calculator or computer to solve this equation. Once you find the solutions, you can check if they are within the given interval [0, 2π).
The answer choices provided are:
a.) π/8
b.) π/8, 5π/8, 9π/8, 13π/8
To determine if these answer choices are correct, you need to substitute each value of x back into the original equation sin(2x) = cos(2x) and see if the equation holds true. If it does, then the answer choice is a solution to the equation.