Find all solution in interval 02pi 2sin2xcos2x2sin2xcos2x1So

Find all solution in interval (0,2pi)

2sin2x+cos2x=2sin2xcos2x+1

Solution

2sin2x + cos2x = 1

2*2sin(x)cos(x) + 1 - 2sin(x)^2 = 1
4sin(x)cos(x) - 2sin(x)^2 = 0
2sin(x)[2cos(x) - sin(x)] = 0

So we can break this up into a system of questions:

{2sin(x) = 0
{2cos(x) - sin(x) = 0
---
{sin(x) = 0
{2cos(x) = sin(x)
---
{sin(x) = 0
{2 = tan(x)

Our first equation is trivial to solve, as sin(x) is equal to 0 for any x of the form n(pi), where n is any integer.

For our second equation, x will equal the arctan(2) + 2n(pi), where n is any integer.