What am I doing wrong?

Equation:
sin2x = 2cos2x

Answers:
90 and 270

....

My Work:

2sin(x)cos(x) = 2cos(2x)
sin(x) cos(x) = cos(2x)
sin(x) cos(x) = 2cos^2(x) - 1
cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1
cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1
5cos^4(x) - 5cos^2(x) + 1 = 0
cos(x) = (+/-) 0.85065080835215
cos(x) = (+/-)0.52573111211905

wouldn't it be easier to solve for 2x then divide it by two?

sin2x=2cos2x

sin2x/cos2x=2
tan 2x=2

2x=63.43 deg
x= .... you do it.

To find the values of x that satisfy the equation sin(2x) = 2cos(2x), you correctly simplified the equation to cos(x) = ±0.85065080835215 and cos(x) = ±0.52573111211905. Now, to solve for x, you can take the inverse cosine (also called arccosine) of both sides of the equation.

Using a scientific calculator or an online tool that can calculate inverse cosine, inputting cos⁻¹(0.85065080835215), you will get x = 30 degrees or x = 150 degrees.

Similarly, inputting cos⁻¹(-0.85065080835215), you will get x = 210 degrees or x = 330 degrees.

Next, inputting cos⁻¹(0.52573111211905), you will get x ≈ 58.28 degrees or x ≈ 92.36 degrees.

And finally, inputting cos⁻¹(-0.52573111211905), you will get x ≈ 67.64 degrees or x ≈ 112.36 degrees.

So, the solutions for x are approximately 30 degrees, 58.28 degrees, 67.64 degrees, 92.36 degrees, 112.36 degrees, 150 degrees, 210 degrees, and 330 degrees.