SOLUTION: what would be the solutions to (sin2x+cos2x)^2=1? I know that this can be expanded to (sin2x+cos2x) (sin2x+cos2x)=1, which can be multiplied to be sin^2(4x)+2sin(2x)cos(2x)+cos^2(

Algebra ->  Trigonometry-basics -> SOLUTION: what would be the solutions to (sin2x+cos2x)^2=1? I know that this can be expanded to (sin2x+cos2x) (sin2x+cos2x)=1, which can be multiplied to be sin^2(4x)+2sin(2x)cos(2x)+cos^2(      Log On


   

Question 1006604: what would be the solutions to (sin2x+cos2x)^2=1?
I know that this can be expanded to (sin2x+cos2x) (sin2x+cos2x)=1, which can be multiplied to be sin^2(4x)+2sin(2x)cos(2x)+cos^2(4x)=1. Or am I missing something here? I'm not sure what to do after this. Please help!
Answer by ikleyn(52812) About Me  (Show Source):
You can put this solution on YOUR website!
.
What would be the solutions to (sin2x + cos2x)^2 = 1?
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Start with 

%28sin2x+%2B+cos2x%29%5E2 = sin%5E2%282x%29 + 2%2Asin%282x%29%2Acos%282x%29 + cos%5E2%282x%29 = 1 + 2*sin(2x)*cos(2x).

Therefore, your equation takes the form

1 + 2*sin(2x)*cos(2x) = 1,   or

2*sin(2x)*cos(2x) = 0.   (1)

Now, do you know this formula sin(2alpha) = 2*sin(alpha)*cos(alpha) ?
(See the lesson Trigonometric functions of multiply argument in this site).

It reduces the equation (1) to

sin(4x) = 0.

The solution is x = k%2A%28pi%2F4%29, k = 0, +/-1, +/-2, . . . 

It is the solution of your original equation.