Question 1006604
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What would be the solutions to (sin2x + cos2x)^2 = 1?
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Start with 

{{{(sin2x + cos2x)^2}}} = {{{sin^2(2x)}}} + {{{2*sin(2x)*cos(2x)}}} + {{{cos^2(2x)}}} = 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 <A HREF =http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-multiply-argument.lesson>Trigonometric functions of multiply argument</A> in this site).

It reduces the equation (1) to

sin(4x) = 0.

The solution is x = {{{k*(pi/4)}}}, k = 0, +/-1, +/-2, . . . 

It is the solution of your original equation.
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