Question

How do you find all solutions to `[sin (2x) + cos (2x)] ^2 = 1`?

Answer 1

Expansion simplifies this to sin (4x ) = 0.
Answer: x = n`pi`/4, n = 0, `+-`1,`+-`2`+-`3,`+-`4,,`+-`5, ...,

Explanation:

4x = n`pi`.